Deck 19: Further Non-Parametric Tests

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Question
The sum of ranks in a Wilcoxon rank sum test must equal _______.

A) n(n + 1)/2
B) n(n - 1)/2
C) n(n - 2).
D) (n * n)/2.
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Question
Instruction 19-1
The director of transportation of a large company is interested in the usage of her van pool. She considers her routes to be divided into local and non-local. She is particularly interested in learning if there is a difference in the proportion of males and females who use the local routes. She takes a sample of a day's riders and finds the following:
 Male  Female  Total  Local 274471 Non-Local 33255 Total 6069129\begin{array} { | l | c | c | l | } \hline & \text { Male } & \text { Female } & \text { Total } \\\hline \text { Local } & 27 & 44 & 71 \\\hline \text { Non-Local } & 33 & 25 & 5 \\\hline \text { Total } & 60 & 69 & 129 \\\hline\end{array} She will use this information to perform a chi-square hypothesis test using a level of significance of 0.05.

-Referring to Instruction 19-1,the director now wants to know if the proportion of users who are male and the proportion of users who use the local routes are the same.Which test should she use?

A) Z-test for difference in proportions
B) Wilcoxon rank sum test
C) McNemar test for difference in proportions
D) (?)2-test for difference in proportions
Question
Given two samples with sizes 20 and 14,respectively,for the Wilcoxon rank sum test the total sample size is _______.

A) 34
B) 35
C) 33
D) 32
Question
The McNemar test is approximately distributed as a standardised normal random variable.
Question
The McNemar test is used to determine whether there is evidence of a difference between the proportions of two related samples.
Question
For the Wilcoxon rank sum test,the quantity T1 is defined as _______.

A) the total sum of ranks minus
B) the sum of ranks in the larger sample
C) the total sum of ranks
D) the sum of ranks of the values in the smaller sample
Question
Instruction 19-1
The director of transportation of a large company is interested in the usage of her van pool. She considers her routes to be divided into local and non-local. She is particularly interested in learning if there is a difference in the proportion of males and females who use the local routes. She takes a sample of a day's riders and finds the following:
 Male  Female  Total  Local 274471 Non-Local 33255 Total 6069129\begin{array} { | l | c | c | l | } \hline & \text { Male } & \text { Female } & \text { Total } \\\hline \text { Local } & 27 & 44 & 71 \\\hline \text { Non-Local } & 33 & 25 & 5 \\\hline \text { Total } & 60 & 69 & 129 \\\hline\end{array} She will use this information to perform a chi-square hypothesis test using a level of significance of 0.05.

-Referring to Instruction 19-1,the director now wants to know if the proportion of users who are male and the proportion of users who use the local routes are the same.What should be her conclusion?

A) There is insufficient evidence that the proportion of users who are male is not the same as the proportion of users who use the local routes.
B) There is sufficient evidence that the proportion of users who are male is not the same as the proportion of users who use the local routes.
C) There is insufficient evidence that the proportion of users who are male is the same as the proportion of users who use the local routes.
D) There is sufficient evidence that the proportion of users who are male is the same as the proportion of users who use the local routes.
Question
One reason for using the Wilcoxon rank sum test rather than the t test is if you cannot assume that the populations from which the samples came are normally distributed.
Question
Instruction 19-1
The director of transportation of a large company is interested in the usage of her van pool. She considers her routes to be divided into local and non-local. She is particularly interested in learning if there is a difference in the proportion of males and females who use the local routes. She takes a sample of a day's riders and finds the following:
 Male  Female  Total  Local 274471 Non-Local 33255 Total 6069129\begin{array} { | l | c | c | l | } \hline & \text { Male } & \text { Female } & \text { Total } \\\hline \text { Local } & 27 & 44 & 71 \\\hline \text { Non-Local } & 33 & 25 & 5 \\\hline \text { Total } & 60 & 69 & 129 \\\hline\end{array} She will use this information to perform a chi-square hypothesis test using a level of significance of 0.05.

-Referring to Instruction 19-1,the director now wants to know if the proportion of users who are female and the proportion of users who use the local routes are the same.Which test should she use?

A) McNemar test for difference in proportions
B) (?)2-test for difference in proportions
C) Z-test for difference in proportions
D) Wilcoxon rank sum test
Question
The McNemar test is approximately chi-square distributed.
Question
Instruction 19-1
The director of transportation of a large company is interested in the usage of her van pool. She considers her routes to be divided into local and non-local. She is particularly interested in learning if there is a difference in the proportion of males and females who use the local routes. She takes a sample of a day's riders and finds the following:
 Male  Female  Total  Local 274471 Non-Local 33255 Total 6069129\begin{array} { | l | c | c | l | } \hline & \text { Male } & \text { Female } & \text { Total } \\\hline \text { Local } & 27 & 44 & 71 \\\hline \text { Non-Local } & 33 & 25 & 5 \\\hline \text { Total } & 60 & 69 & 129 \\\hline\end{array} She will use this information to perform a chi-square hypothesis test using a level of significance of 0.05.

-Referring to Instruction 19-1,the director now wants to know if the proportion of users who are male and the proportion of users who use the local routes are the same.What is the p-value of the test statistic using a 5% level of significance?
Question
Instruction 19-1
The director of transportation of a large company is interested in the usage of her van pool. She considers her routes to be divided into local and non-local. She is particularly interested in learning if there is a difference in the proportion of males and females who use the local routes. She takes a sample of a day's riders and finds the following:
 Male  Female  Total  Local 274471 Non-Local 33255 Total 6069129\begin{array} { | l | c | c | l | } \hline & \text { Male } & \text { Female } & \text { Total } \\\hline \text { Local } & 27 & 44 & 71 \\\hline \text { Non-Local } & 33 & 25 & 5 \\\hline \text { Total } & 60 & 69 & 129 \\\hline\end{array} She will use this information to perform a chi-square hypothesis test using a level of significance of 0.05.

-Referring to Instruction 19-1,the director now wants to know if the proportion of users who are male and the proportion of users who use the local routes are the same.She should reject the null hypothesis using a 5% level of significance.
Question
The McNemar test is approximately distributed as a chi-square random variable.
Question
One disadvantage of the Wilcoxon rank sum test,as compared to the t test,is that Wilcoxon rank sum test is much less powerful.
Question
Instruction 19-1
The director of transportation of a large company is interested in the usage of her van pool. She considers her routes to be divided into local and non-local. She is particularly interested in learning if there is a difference in the proportion of males and females who use the local routes. She takes a sample of a day's riders and finds the following:
 Male  Female  Total  Local 274471 Non-Local 33255 Total 6069129\begin{array} { | l | c | c | l | } \hline & \text { Male } & \text { Female } & \text { Total } \\\hline \text { Local } & 27 & 44 & 71 \\\hline \text { Non-Local } & 33 & 25 & 5 \\\hline \text { Total } & 60 & 69 & 129 \\\hline\end{array} She will use this information to perform a chi-square hypothesis test using a level of significance of 0.05.

-Referring to Instruction 19-1,the director now wants to know if the proportion of users who are male and the proportion of users who use the local routes are the same.She should reject the null hypothesis using a 5% level of significance.
Question
The Wilcoxon rank sum test is a non-parametric test for testing the difference between two ____from independent samples.
Question
Instruction 19-1
The director of transportation of a large company is interested in the usage of her van pool. She considers her routes to be divided into local and non-local. She is particularly interested in learning if there is a difference in the proportion of males and females who use the local routes. She takes a sample of a day's riders and finds the following:
 Male  Female  Total  Local 274471 Non-Local 33255 Total 6069129\begin{array} { | l | c | c | l | } \hline & \text { Male } & \text { Female } & \text { Total } \\\hline \text { Local } & 27 & 44 & 71 \\\hline \text { Non-Local } & 33 & 25 & 5 \\\hline \text { Total } & 60 & 69 & 129 \\\hline\end{array} She will use this information to perform a chi-square hypothesis test using a level of significance of 0.05.

-Referring to Instruction 19-1,the director now wants to know if the proportion of users who are female and the proportion of users who use the local routes are the same.What is the value of the test statistic using a 5% level of significance?
Question
Instruction 19-1
The director of transportation of a large company is interested in the usage of her van pool. She considers her routes to be divided into local and non-local. She is particularly interested in learning if there is a difference in the proportion of males and females who use the local routes. She takes a sample of a day's riders and finds the following:
 Male  Female  Total  Local 274471 Non-Local 33255 Total 6069129\begin{array} { | l | c | c | l | } \hline & \text { Male } & \text { Female } & \text { Total } \\\hline \text { Local } & 27 & 44 & 71 \\\hline \text { Non-Local } & 33 & 25 & 5 \\\hline \text { Total } & 60 & 69 & 129 \\\hline\end{array} She will use this information to perform a chi-square hypothesis test using a level of significance of 0.05.

-Referring to Instruction 19-1,the director now wants to know if the proportion of users who are male and the proportion of users who use the local routes are the same.What is the p-value of the test statistic using a 5% level of significance?
Question
Instruction 19-2
The director of a university's MBA program wanted to know if a one week orientation would change the proportion among potential incoming students who would perceive the program as being good. Given below is the result from 215 students' view of the program before and after the orientation.
After the Orientation
 Before the Orientation  Good  Not Good  Total  Good 9337130 Not Good 711485 Total 16451215\begin{array} { | c | c | c | c | } \hline \text { Before the Orientation } & \text { Good } & \text { Not Good } & \text { Total } \\\hline \text { Good } & 93 & 37 & 130 \\\hline \text { Not Good } & 71 & 14 & 85 \\\hline \text { Total } & 164 & 51 & 215 \\\hline\end{array}

-Referring to Instruction 19-2,the director should reject the null hypothesis using a 1% level of significance.
Question
Instruction 19-1
The director of transportation of a large company is interested in the usage of her van pool. She considers her routes to be divided into local and non-local. She is particularly interested in learning if there is a difference in the proportion of males and females who use the local routes. She takes a sample of a day's riders and finds the following:
 Male  Female  Total  Local 274471 Non-Local 33255 Total 6069129\begin{array} { | l | c | c | l | } \hline & \text { Male } & \text { Female } & \text { Total } \\\hline \text { Local } & 27 & 44 & 71 \\\hline \text { Non-Local } & 33 & 25 & 5 \\\hline \text { Total } & 60 & 69 & 129 \\\hline\end{array} She will use this information to perform a chi-square hypothesis test using a level of significance of 0.05.

-Referring to Instruction 19-1,the director now wants to know if the proportion of users who are male and the proportion of users who use the local routes are the same.What is the value of the test statistic using a 5% level of significance?
Question
Instruction 19-3
A perfume manufacturer is trying to choose between two magazine advertising layouts. An expensive layout would include a small package of the perfume. A cheaper layout would include a 'scratch-and-sniff' sample of the product. The manufacturer would use the more expensive layout only if there is evidence that it would lead to a higher approval rate. The manufacturer presents both layouts to five groups and determines the approval rating from each group on both layouts. The data are given below. Use this to test whether the median difference in approval rating is different from zero in favour of the more expensive layout with a level of significance of 0.05.
 Padkage  Scratch 52376843435348395647\begin{array}{cc}\text { Padkage } & \text { Scratch } \\\hline 52 & 37 \\68 & 43 \\43 & 53 \\48 & 39 \\56 & 47\end{array}


-Referring to Instruction 19-3,the perfume manufacturer will ____.

A) use the package layout because there is sufficient evidence to conclude that this is the best course of action
B) use the "scratch-and-sniff" layout because there is insufficient evidence to do otherwise
C) use the "scratch-and-sniff" layout because there is sufficient evidence to conclude that this is the best course of action
D) use the package layout because there is insufficient evidence to do otherwise
Question
Instruction 19-5
The director of the MBA program of a university wanted to know if a one week orientation would change the proportion among potential incoming students who would perceive the program as being good. Given below is the result from 215 students' view of the program before and after the orientation.
After the Orientation
 Before the Orientation  Good  Not Good  Total  Good 9337130 Not Good 711485 Total 16451215\begin{array} { | c | c | c | c | } \hline \text { Before the Orientation } & \text { Good } & \text { Not Good } & \text { Total } \\\hline \text { Good } & 93 & 37 & 130 \\\hline \text { Not Good } & 71 & 14 & 85 \\\hline \text { Total } & 164 & 51 & 215 \\\hline\end{array}

-Referring to Instruction 19-5,the null hypothesis for the Friedman rank test for the difference in the means should be rejected at a 0.01 level of significance.
Question
If the assumptions of the one-way ANOVA F test are not met,which of the following test could be used?

A) Friedman rank test
B) Marascuilo procedure
C) Kruskal-Wallis test
D) McNemar test
Question
The Kruskal-Wallis rank test for differences in more than two medians is a nonparametric alternative to ____

A) Student's t test for independent samples
B) Wilcoxon's rank sum test for differences in two medians
C) ANOVA F test for completely randomised experiments
D) Student's t test for related samples
Question
Instruction 19-5
The director of the MBA program of a university wanted to know if a one week orientation would change the proportion among potential incoming students who would perceive the program as being good. Given below is the result from 215 students' view of the program before and after the orientation.
After the Orientation
 Before the Orientation  Good  Not Good  Total  Good 9337130 Not Good 711485 Total 16451215\begin{array} { | c | c | c | c | } \hline \text { Before the Orientation } & \text { Good } & \text { Not Good } & \text { Total } \\\hline \text { Good } & 93 & 37 & 130 \\\hline \text { Not Good } & 71 & 14 & 85 \\\hline \text { Total } & 164 & 51 & 215 \\\hline\end{array}

-Referring to Instruction 19-5,the decision made at a 0.01 level of significance on the Friedman rank test for the difference in medians implies that the three medians are not all the same.
Question
Instruction 19-5
The director of the MBA program of a university wanted to know if a one week orientation would change the proportion among potential incoming students who would perceive the program as being good. Given below is the result from 215 students' view of the program before and after the orientation.
After the Orientation
 Before the Orientation  Good  Not Good  Total  Good 9337130 Not Good 711485 Total 16451215\begin{array} { | c | c | c | c | } \hline \text { Before the Orientation } & \text { Good } & \text { Not Good } & \text { Total } \\\hline \text { Good } & 93 & 37 & 130 \\\hline \text { Not Good } & 71 & 14 & 85 \\\hline \text { Total } & 164 & 51 & 215 \\\hline\end{array}

-Referring to Instruction 19-5,the decision made at a 0.01 level of significance on the Friedman rank test for the difference in medians implies that all three medians are significantly different.
Question
Instruction 19-5
The director of the MBA program of a university wanted to know if a one week orientation would change the proportion among potential incoming students who would perceive the program as being good. Given below is the result from 215 students' view of the program before and after the orientation.
After the Orientation
 Before the Orientation  Good  Not Good  Total  Good 9337130 Not Good 711485 Total 16451215\begin{array} { | c | c | c | c | } \hline \text { Before the Orientation } & \text { Good } & \text { Not Good } & \text { Total } \\\hline \text { Good } & 93 & 37 & 130 \\\hline \text { Not Good } & 71 & 14 & 85 \\\hline \text { Total } & 164 & 51 & 215 \\\hline\end{array}

-Referring to Instruction 19-5,the Friedman rank test is valid only if there is no interaction between the five blocks and the three treatment levels.
Question
Instruction 19-3
A perfume manufacturer is trying to choose between two magazine advertising layouts. An expensive layout would include a small package of the perfume. A cheaper layout would include a 'scratch-and-sniff' sample of the product. The manufacturer would use the more expensive layout only if there is evidence that it would lead to a higher approval rate. The manufacturer presents both layouts to five groups and determines the approval rating from each group on both layouts. The data are given below. Use this to test whether the median difference in approval rating is different from zero in favour of the more expensive layout with a level of significance of 0.05.
 Padkage  Scratch 52376843435348395647\begin{array}{cc}\text { Padkage } & \text { Scratch } \\\hline 52 & 37 \\68 & 43 \\43 & 53 \\48 & 39 \\56 & 47\end{array}


-Referring to Instruction 19-3,what is the rank of the absolute difference for the last pair of observations?
Question
The Kruskal-Wallis test is an extension of which of the following for two independent samples?

A) Paired-sample t test
B) Wilcoxon rank sum test
C) Pooled-variance t test
D) Wilcoxon signed ranks test
Question
Instruction 19-3
A perfume manufacturer is trying to choose between two magazine advertising layouts. An expensive layout would include a small package of the perfume. A cheaper layout would include a 'scratch-and-sniff' sample of the product. The manufacturer would use the more expensive layout only if there is evidence that it would lead to a higher approval rate. The manufacturer presents both layouts to five groups and determines the approval rating from each group on both layouts. The data are given below. Use this to test whether the median difference in approval rating is different from zero in favour of the more expensive layout with a level of significance of 0.05.
 Padkage  Scratch 52376843435348395647\begin{array}{cc}\text { Padkage } & \text { Scratch } \\\hline 52 & 37 \\68 & 43 \\43 & 53 \\48 & 39 \\56 & 47\end{array}


-Referring to Instruction 19-3,what are the lower and upper critical values of the test?
Question
Instruction 19-5
The director of the MBA program of a university wanted to know if a one week orientation would change the proportion among potential incoming students who would perceive the program as being good. Given below is the result from 215 students' view of the program before and after the orientation.
After the Orientation
 Before the Orientation  Good  Not Good  Total  Good 9337130 Not Good 711485 Total 16451215\begin{array} { | c | c | c | c | } \hline \text { Before the Orientation } & \text { Good } & \text { Not Good } & \text { Total } \\\hline \text { Good } & 93 & 37 & 130 \\\hline \text { Not Good } & 71 & 14 & 85 \\\hline \text { Total } & 164 & 51 & 215 \\\hline\end{array}

-Referring to Instruction 19-5,the Friedman rank test is valid only if the five blocks are independent so that the yields in one block have no influence on the yields in any other block.
Question
The Journal of Business Venturing reported on the activities of entrepreneurs during the organisation creation process.As part of a designed study,a total of 71 entrepreneurs were interviewed and divided into three groups: those that were successful in founding a new firm (n1 = 34),those still actively trying to establish a firm (n2 = 21),and those who tried to start a new firm but eventually gave up (n3 = 16).The total number of activities undertaken (e.g.,developed a business plan,sought funding,looked for facilities)by each group over a specified time period during organisation creation was measured.The objective is to compare the mean number of activities of the three groups of entrepreneurs.Because of concerns over necessary assumption of the parametric analysis,it was decided to use a nonparametric analysis.Identify the nonparametric method that would be used to analyse the data.

A) Kruskal-Wallis rank test for differences in medians
B) Wilcoxon signed rank test
C) Wilcoxon rank sum test
D) One-way ANOVA F test
Question
Instruction 19-3
A perfume manufacturer is trying to choose between two magazine advertising layouts. An expensive layout would include a small package of the perfume. A cheaper layout would include a 'scratch-and-sniff' sample of the product. The manufacturer would use the more expensive layout only if there is evidence that it would lead to a higher approval rate. The manufacturer presents both layouts to five groups and determines the approval rating from each group on both layouts. The data are given below. Use this to test whether the median difference in approval rating is different from zero in favour of the more expensive layout with a level of significance of 0.05.
 Padkage  Scratch 52376843435348395647\begin{array}{cc}\text { Padkage } & \text { Scratch } \\\hline 52 & 37 \\68 & 43 \\43 & 53 \\48 & 39 \\56 & 47\end{array}


-Referring to Instruction 19-3,the hypotheses that should be used are

A) H0: MD = 0 versus H1: MD ≠\neq 0.
B) H0: M1 ≠\neq 0 versus H1: M1 ≠\neq M2.
C)H0: MD ≤\le 0 versus H1: MD > 0.
D)H0: M1 ≤\le M2 versus H1: M1 > M2.
Question
Instruction 19-3
A perfume manufacturer is trying to choose between two magazine advertising layouts. An expensive layout would include a small package of the perfume. A cheaper layout would include a 'scratch-and-sniff' sample of the product. The manufacturer would use the more expensive layout only if there is evidence that it would lead to a higher approval rate. The manufacturer presents both layouts to five groups and determines the approval rating from each group on both layouts. The data are given below. Use this to test whether the median difference in approval rating is different from zero in favour of the more expensive layout with a level of significance of 0.05.
 Padkage  Scratch 52376843435348395647\begin{array}{cc}\text { Padkage } & \text { Scratch } \\\hline 52 & 37 \\68 & 43 \\43 & 53 \\48 & 39 \\56 & 47\end{array}


-Referring to Instruction 19-3,what is the value of the test statistic?
Question
Instruction 19-4
An agronomist wants to compare the crop yield of three varieties of chickpea seeds. She plants all three varieties of the seeds on each of five different patches of fields. She then measures the crop yield in bushels per acre. Treating this as a randomised block design, the results are presented in the table that follows.
 Fields  Smith  Walsh  Trevor 111.119.014.6213.518.015.7315.319.816.8414.619.616.759.816.615.2\begin{array} { | c | c | c | c | } \hline \text { Fields } & \text { Smith } & \text { Walsh } & \text { Trevor } \\\hline 1 & 11.1 & 19.0 & 14.6 \\\hline 2 & 13.5 & 18.0 & 15.7 \\\hline 3 & 15.3 & 19.8 & 16.8 \\\hline 4 & 14.6 & 19.6 & 16.7 \\\hline 5 & 9.8 & 16.6 & 15.2 \\\hline\end{array} Below is the Minitab output of the Friedman rank test:
Friectman Test Yedd versus Varieties Felds
Friedmantest for Yield by Varieties blocked by Fields S=10.00DF=2P=0.007S = 10.00 D F = 2 P = 0.007

 Varieties N EstMedian  Sum of Ranks  Smith 513.5005.0 Trevor 515.66710.0 Walsh 518.53315.0\begin{array} { l l l l } \text { Varieties } & N & \text { EstMedian } & \text { Sum of Ranks } \\ & & & \\ \text { Smith } & 5 & 13.500 & 5.0 \\ \text { Trevor } & 5 & 15.667 & 10.0 \\ \text { Walsh } & 5 & 18.533 & 15.0 \end{array}

 Grand median =15.900\text { Grand median }=15.900


-Referring to Instruction 19-4,the null hypothesis for the Friedman rank test is

A) MSmith = MWalsh = MTrevor.
B)H0: M Field 1 = M Field 2 = MField 3 = MField 4 = MField 5.
C)H0: μ\mu Field 1 = μ\mu Field 2 = μ\mu Field 3 = μ\mu Field 4 = μ\mu Field 5.
D) H0: μ\mu Smith = μ\mu Walsh = μ\mu Trevor.
Question
Suppose there is interest in comparing the median response time for three independent groups learning a specific task.The appropriate nonparametric procedure is ____.

A) Wilcoxon signed rank test
B) Kruskal-Wallis rank test for differences in medians
C) Wilcoxon rank sum test
D) Friedman rank test
Question
If the sample sizes in each group are more than five,the Kruskal-Wallis rank test statistic can be approximated by a standardised normal distribution.
Question
If the sample sizes in each group are more than five,the Kruskal-Wallis rank test statistic can be approximated by a chi-square distribution.
Question
Instruction 19-3
A perfume manufacturer is trying to choose between two magazine advertising layouts. An expensive layout would include a small package of the perfume. A cheaper layout would include a 'scratch-and-sniff' sample of the product. The manufacturer would use the more expensive layout only if there is evidence that it would lead to a higher approval rate. The manufacturer presents both layouts to five groups and determines the approval rating from each group on both layouts. The data are given below. Use this to test whether the median difference in approval rating is different from zero in favour of the more expensive layout with a level of significance of 0.05.
 Padkage  Scratch 52376843435348395647\begin{array}{cc}\text { Padkage } & \text { Scratch } \\\hline 52 & 37 \\68 & 43 \\43 & 53 \\48 & 39 \\56 & 47\end{array}


-Referring to Instruction 19-3,which pair(s)of observations has a negative signed rank?
Question
Instruction 19-3
A perfume manufacturer is trying to choose between two magazine advertising layouts. An expensive layout would include a small package of the perfume. A cheaper layout would include a 'scratch-and-sniff' sample of the product. The manufacturer would use the more expensive layout only if there is evidence that it would lead to a higher approval rate. The manufacturer presents both layouts to five groups and determines the approval rating from each group on both layouts. The data are given below. Use this to test whether the median difference in approval rating is different from zero in favour of the more expensive layout with a level of significance of 0.05.
 Padkage  Scratch 52376843435348395647\begin{array}{cc}\text { Padkage } & \text { Scratch } \\\hline 52 & 37 \\68 & 43 \\43 & 53 \\48 & 39 \\56 & 47\end{array}


-Referring to Instruction 19-3,what is the appropriate test to use?

A) Wilcoxon signed rank test for difference in median
B) Wilcoxon signed rank test for median difference
C) Wilcoxon rank sum test for median difference
D) Wilcoxon rank sum test for difference in median
Question
If the number of blocks in the experiment is more than five,the Friedman rank test statistic can be approximated by a standardised normal distribution.
Question
A non-parametric test for testing the difference between two medians from independent samples is known as the ____.

A) Kruskal-Wallis test
B) Wilcoxon rank sum test
C) Friedman rank test
D) Wilcoxon signed ranks test
Question
The ____ finds whether multiple sample groups have been selected from populations with equal medians.

A) Kruskal-Wallis test
B) Wilcoxon rank sum test
C) Friedman rank test
D) Wilcoxon signed ranks test
Question
If the number of blocks in the experiment is more than five,the Friedman rank test statistic can be approximated by a chi-square distribution.
Question
Instruction 19-6
An agronomist wants to compare the crop yield of three varieties of chickpea seeds. She plants all three varieties of the seeds on each of five different patches of fields. She then measures the crop yield in bushels per acre. Treating this as a randomised block design, the results are presented in the table that follows.
 Fields  Smith  Walsh  Trevor 111.119.014.6213.518.015.7315.319.816.8414.619.616.759.816.615.2\begin{array} { | c | c | c | c | } \hline \text { Fields } & \text { Smith } & \text { Walsh } & \text { Trevor } \\\hline 1 & 11.1 & 19.0 & 14.6 \\\hline2 & 13.5 & 18.0 & 15.7 \\\hline 3 & 15.3 & 19.8 & 16.8 \\\hline 4 & 14.6 & 19.6 & 16.7 \\\hline 5& 9.8 & 16.6 & 15.2 \\\hline\end{array} Below is the Minitab output of the Friedman rank test:
Friectman Test Yedd versus Varieties Felds
Friedmantest for Yield by Varieties blocked by Fields S=10.00DF=2P=0.007S = 10.00 D F = 2 P = 0.007

 Varieties N EstMedian  Sum of Ranks  Smith 513.5005.0 Trevor 515.66710.0 Walsh 518.53315.0\begin{array} { l l l l } \text { Varieties } & N & \text { EstMedian } & \text { Sum of Ranks } \\ & & & \\ \text { Smith } & 5 & 13.500 & 5.0 \\ \text { Trevor } & 5 & 15.667 & 10.0 \\ \text { Walsh } & 5 & 18.533 & 15.0 \end{array}

 Grand median =15.900\text { Grand median }=15.900


-Referring to Instruction 19-6,what is the value of the test statistic for the Friedman rank test for the difference in the medians?
Question
Instruction 19-6
An agronomist wants to compare the crop yield of three varieties of chickpea seeds. She plants all three varieties of the seeds on each of five different patches of fields. She then measures the crop yield in bushels per acre. Treating this as a randomised block design, the results are presented in the table that follows.
 Fields  Smith  Walsh  Trevor 111.119.014.6213.518.015.7315.319.816.8414.619.616.759.816.615.2\begin{array} { | c | c | c | c | } \hline \text { Fields } & \text { Smith } & \text { Walsh } & \text { Trevor } \\\hline 1 & 11.1 & 19.0 & 14.6 \\\hline2 & 13.5 & 18.0 & 15.7 \\\hline 3 & 15.3 & 19.8 & 16.8 \\\hline 4 & 14.6 & 19.6 & 16.7 \\\hline 5& 9.8 & 16.6 & 15.2 \\\hline\end{array} Below is the Minitab output of the Friedman rank test:
Friectman Test Yedd versus Varieties Felds
Friedmantest for Yield by Varieties blocked by Fields S=10.00DF=2P=0.007S = 10.00 D F = 2 P = 0.007

 Varieties N EstMedian  Sum of Ranks  Smith 513.5005.0 Trevor 515.66710.0 Walsh 518.53315.0\begin{array} { l l l l } \text { Varieties } & N & \text { EstMedian } & \text { Sum of Ranks } \\ & & & \\ \text { Smith } & 5 & 13.500 & 5.0 \\ \text { Trevor } & 5 & 15.667 & 10.0 \\ \text { Walsh } & 5 & 18.533 & 15.0 \end{array}

 Grand median =15.900\text { Grand median }=15.900


-Referring to Instruction 19-6,what are the degrees of freedom of the Friedman rank test for the difference in the medians at a level of significance of 0.01?
Question
When the normality assumption is not met in a randomised block design,which of the following tests should be used?

A) Kruskal-Wallis test
B) Wilcoxon rank sum test
C) Friedman rank test
D) Wilcoxon signed ranks test
Question
A non-parametric test for testing the median difference for paired samples is known as the ____.

A) Kruskal-Wallis test
B) Wilcoxon rank sum test
C) Friedman rank test
D) Wilcoxon signed ranks test
Question
A non-parametric test for testing the difference between two proportions from related samples is known as the ____.

A) Kruskal-Wallis test
B) Wilcoxon rank sum test
C) McNemar test
D) Wilcoxon signed ranks test
Question
Instruction 19-6
An agronomist wants to compare the crop yield of three varieties of chickpea seeds. She plants all three varieties of the seeds on each of five different patches of fields. She then measures the crop yield in bushels per acre. Treating this as a randomised block design, the results are presented in the table that follows.
 Fields  Smith  Walsh  Trevor 111.119.014.6213.518.015.7315.319.816.8414.619.616.759.816.615.2\begin{array} { | c | c | c | c | } \hline \text { Fields } & \text { Smith } & \text { Walsh } & \text { Trevor } \\\hline 1 & 11.1 & 19.0 & 14.6 \\\hline2 & 13.5 & 18.0 & 15.7 \\\hline 3 & 15.3 & 19.8 & 16.8 \\\hline 4 & 14.6 & 19.6 & 16.7 \\\hline 5& 9.8 & 16.6 & 15.2 \\\hline\end{array} Below is the Minitab output of the Friedman rank test:
Friectman Test Yedd versus Varieties Felds
Friedmantest for Yield by Varieties blocked by Fields S=10.00DF=2P=0.007S = 10.00 D F = 2 P = 0.007

 Varieties N EstMedian  Sum of Ranks  Smith 513.5005.0 Trevor 515.66710.0 Walsh 518.53315.0\begin{array} { l l l l } \text { Varieties } & N & \text { EstMedian } & \text { Sum of Ranks } \\ & & & \\ \text { Smith } & 5 & 13.500 & 5.0 \\ \text { Trevor } & 5 & 15.667 & 10.0 \\ \text { Walsh } & 5 & 18.533 & 15.0 \end{array}

 Grand median =15.900\text { Grand median }=15.900


-Referring to Instruction 19-6,what is the critical value of the Friedman rank test for the difference in the medians at a level of significance of 0.01?
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Deck 19: Further Non-Parametric Tests
1
The sum of ranks in a Wilcoxon rank sum test must equal _______.

A) n(n + 1)/2
B) n(n - 1)/2
C) n(n - 2).
D) (n * n)/2.
n(n + 1)/2
2
Instruction 19-1
The director of transportation of a large company is interested in the usage of her van pool. She considers her routes to be divided into local and non-local. She is particularly interested in learning if there is a difference in the proportion of males and females who use the local routes. She takes a sample of a day's riders and finds the following:
 Male  Female  Total  Local 274471 Non-Local 33255 Total 6069129\begin{array} { | l | c | c | l | } \hline & \text { Male } & \text { Female } & \text { Total } \\\hline \text { Local } & 27 & 44 & 71 \\\hline \text { Non-Local } & 33 & 25 & 5 \\\hline \text { Total } & 60 & 69 & 129 \\\hline\end{array} She will use this information to perform a chi-square hypothesis test using a level of significance of 0.05.

-Referring to Instruction 19-1,the director now wants to know if the proportion of users who are male and the proportion of users who use the local routes are the same.Which test should she use?

A) Z-test for difference in proportions
B) Wilcoxon rank sum test
C) McNemar test for difference in proportions
D) (?)2-test for difference in proportions
McNemar test for difference in proportions
3
Given two samples with sizes 20 and 14,respectively,for the Wilcoxon rank sum test the total sample size is _______.

A) 34
B) 35
C) 33
D) 32
34
4
The McNemar test is approximately distributed as a standardised normal random variable.
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5
The McNemar test is used to determine whether there is evidence of a difference between the proportions of two related samples.
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6
For the Wilcoxon rank sum test,the quantity T1 is defined as _______.

A) the total sum of ranks minus
B) the sum of ranks in the larger sample
C) the total sum of ranks
D) the sum of ranks of the values in the smaller sample
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7
Instruction 19-1
The director of transportation of a large company is interested in the usage of her van pool. She considers her routes to be divided into local and non-local. She is particularly interested in learning if there is a difference in the proportion of males and females who use the local routes. She takes a sample of a day's riders and finds the following:
 Male  Female  Total  Local 274471 Non-Local 33255 Total 6069129\begin{array} { | l | c | c | l | } \hline & \text { Male } & \text { Female } & \text { Total } \\\hline \text { Local } & 27 & 44 & 71 \\\hline \text { Non-Local } & 33 & 25 & 5 \\\hline \text { Total } & 60 & 69 & 129 \\\hline\end{array} She will use this information to perform a chi-square hypothesis test using a level of significance of 0.05.

-Referring to Instruction 19-1,the director now wants to know if the proportion of users who are male and the proportion of users who use the local routes are the same.What should be her conclusion?

A) There is insufficient evidence that the proportion of users who are male is not the same as the proportion of users who use the local routes.
B) There is sufficient evidence that the proportion of users who are male is not the same as the proportion of users who use the local routes.
C) There is insufficient evidence that the proportion of users who are male is the same as the proportion of users who use the local routes.
D) There is sufficient evidence that the proportion of users who are male is the same as the proportion of users who use the local routes.
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8
One reason for using the Wilcoxon rank sum test rather than the t test is if you cannot assume that the populations from which the samples came are normally distributed.
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9
Instruction 19-1
The director of transportation of a large company is interested in the usage of her van pool. She considers her routes to be divided into local and non-local. She is particularly interested in learning if there is a difference in the proportion of males and females who use the local routes. She takes a sample of a day's riders and finds the following:
 Male  Female  Total  Local 274471 Non-Local 33255 Total 6069129\begin{array} { | l | c | c | l | } \hline & \text { Male } & \text { Female } & \text { Total } \\\hline \text { Local } & 27 & 44 & 71 \\\hline \text { Non-Local } & 33 & 25 & 5 \\\hline \text { Total } & 60 & 69 & 129 \\\hline\end{array} She will use this information to perform a chi-square hypothesis test using a level of significance of 0.05.

-Referring to Instruction 19-1,the director now wants to know if the proportion of users who are female and the proportion of users who use the local routes are the same.Which test should she use?

A) McNemar test for difference in proportions
B) (?)2-test for difference in proportions
C) Z-test for difference in proportions
D) Wilcoxon rank sum test
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10
The McNemar test is approximately chi-square distributed.
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11
Instruction 19-1
The director of transportation of a large company is interested in the usage of her van pool. She considers her routes to be divided into local and non-local. She is particularly interested in learning if there is a difference in the proportion of males and females who use the local routes. She takes a sample of a day's riders and finds the following:
 Male  Female  Total  Local 274471 Non-Local 33255 Total 6069129\begin{array} { | l | c | c | l | } \hline & \text { Male } & \text { Female } & \text { Total } \\\hline \text { Local } & 27 & 44 & 71 \\\hline \text { Non-Local } & 33 & 25 & 5 \\\hline \text { Total } & 60 & 69 & 129 \\\hline\end{array} She will use this information to perform a chi-square hypothesis test using a level of significance of 0.05.

-Referring to Instruction 19-1,the director now wants to know if the proportion of users who are male and the proportion of users who use the local routes are the same.What is the p-value of the test statistic using a 5% level of significance?
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12
Instruction 19-1
The director of transportation of a large company is interested in the usage of her van pool. She considers her routes to be divided into local and non-local. She is particularly interested in learning if there is a difference in the proportion of males and females who use the local routes. She takes a sample of a day's riders and finds the following:
 Male  Female  Total  Local 274471 Non-Local 33255 Total 6069129\begin{array} { | l | c | c | l | } \hline & \text { Male } & \text { Female } & \text { Total } \\\hline \text { Local } & 27 & 44 & 71 \\\hline \text { Non-Local } & 33 & 25 & 5 \\\hline \text { Total } & 60 & 69 & 129 \\\hline\end{array} She will use this information to perform a chi-square hypothesis test using a level of significance of 0.05.

-Referring to Instruction 19-1,the director now wants to know if the proportion of users who are male and the proportion of users who use the local routes are the same.She should reject the null hypothesis using a 5% level of significance.
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13
The McNemar test is approximately distributed as a chi-square random variable.
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14
One disadvantage of the Wilcoxon rank sum test,as compared to the t test,is that Wilcoxon rank sum test is much less powerful.
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15
Instruction 19-1
The director of transportation of a large company is interested in the usage of her van pool. She considers her routes to be divided into local and non-local. She is particularly interested in learning if there is a difference in the proportion of males and females who use the local routes. She takes a sample of a day's riders and finds the following:
 Male  Female  Total  Local 274471 Non-Local 33255 Total 6069129\begin{array} { | l | c | c | l | } \hline & \text { Male } & \text { Female } & \text { Total } \\\hline \text { Local } & 27 & 44 & 71 \\\hline \text { Non-Local } & 33 & 25 & 5 \\\hline \text { Total } & 60 & 69 & 129 \\\hline\end{array} She will use this information to perform a chi-square hypothesis test using a level of significance of 0.05.

-Referring to Instruction 19-1,the director now wants to know if the proportion of users who are male and the proportion of users who use the local routes are the same.She should reject the null hypothesis using a 5% level of significance.
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16
The Wilcoxon rank sum test is a non-parametric test for testing the difference between two ____from independent samples.
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17
Instruction 19-1
The director of transportation of a large company is interested in the usage of her van pool. She considers her routes to be divided into local and non-local. She is particularly interested in learning if there is a difference in the proportion of males and females who use the local routes. She takes a sample of a day's riders and finds the following:
 Male  Female  Total  Local 274471 Non-Local 33255 Total 6069129\begin{array} { | l | c | c | l | } \hline & \text { Male } & \text { Female } & \text { Total } \\\hline \text { Local } & 27 & 44 & 71 \\\hline \text { Non-Local } & 33 & 25 & 5 \\\hline \text { Total } & 60 & 69 & 129 \\\hline\end{array} She will use this information to perform a chi-square hypothesis test using a level of significance of 0.05.

-Referring to Instruction 19-1,the director now wants to know if the proportion of users who are female and the proportion of users who use the local routes are the same.What is the value of the test statistic using a 5% level of significance?
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18
Instruction 19-1
The director of transportation of a large company is interested in the usage of her van pool. She considers her routes to be divided into local and non-local. She is particularly interested in learning if there is a difference in the proportion of males and females who use the local routes. She takes a sample of a day's riders and finds the following:
 Male  Female  Total  Local 274471 Non-Local 33255 Total 6069129\begin{array} { | l | c | c | l | } \hline & \text { Male } & \text { Female } & \text { Total } \\\hline \text { Local } & 27 & 44 & 71 \\\hline \text { Non-Local } & 33 & 25 & 5 \\\hline \text { Total } & 60 & 69 & 129 \\\hline\end{array} She will use this information to perform a chi-square hypothesis test using a level of significance of 0.05.

-Referring to Instruction 19-1,the director now wants to know if the proportion of users who are male and the proportion of users who use the local routes are the same.What is the p-value of the test statistic using a 5% level of significance?
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19
Instruction 19-2
The director of a university's MBA program wanted to know if a one week orientation would change the proportion among potential incoming students who would perceive the program as being good. Given below is the result from 215 students' view of the program before and after the orientation.
After the Orientation
 Before the Orientation  Good  Not Good  Total  Good 9337130 Not Good 711485 Total 16451215\begin{array} { | c | c | c | c | } \hline \text { Before the Orientation } & \text { Good } & \text { Not Good } & \text { Total } \\\hline \text { Good } & 93 & 37 & 130 \\\hline \text { Not Good } & 71 & 14 & 85 \\\hline \text { Total } & 164 & 51 & 215 \\\hline\end{array}

-Referring to Instruction 19-2,the director should reject the null hypothesis using a 1% level of significance.
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20
Instruction 19-1
The director of transportation of a large company is interested in the usage of her van pool. She considers her routes to be divided into local and non-local. She is particularly interested in learning if there is a difference in the proportion of males and females who use the local routes. She takes a sample of a day's riders and finds the following:
 Male  Female  Total  Local 274471 Non-Local 33255 Total 6069129\begin{array} { | l | c | c | l | } \hline & \text { Male } & \text { Female } & \text { Total } \\\hline \text { Local } & 27 & 44 & 71 \\\hline \text { Non-Local } & 33 & 25 & 5 \\\hline \text { Total } & 60 & 69 & 129 \\\hline\end{array} She will use this information to perform a chi-square hypothesis test using a level of significance of 0.05.

-Referring to Instruction 19-1,the director now wants to know if the proportion of users who are male and the proportion of users who use the local routes are the same.What is the value of the test statistic using a 5% level of significance?
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21
Instruction 19-3
A perfume manufacturer is trying to choose between two magazine advertising layouts. An expensive layout would include a small package of the perfume. A cheaper layout would include a 'scratch-and-sniff' sample of the product. The manufacturer would use the more expensive layout only if there is evidence that it would lead to a higher approval rate. The manufacturer presents both layouts to five groups and determines the approval rating from each group on both layouts. The data are given below. Use this to test whether the median difference in approval rating is different from zero in favour of the more expensive layout with a level of significance of 0.05.
 Padkage  Scratch 52376843435348395647\begin{array}{cc}\text { Padkage } & \text { Scratch } \\\hline 52 & 37 \\68 & 43 \\43 & 53 \\48 & 39 \\56 & 47\end{array}


-Referring to Instruction 19-3,the perfume manufacturer will ____.

A) use the package layout because there is sufficient evidence to conclude that this is the best course of action
B) use the "scratch-and-sniff" layout because there is insufficient evidence to do otherwise
C) use the "scratch-and-sniff" layout because there is sufficient evidence to conclude that this is the best course of action
D) use the package layout because there is insufficient evidence to do otherwise
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22
Instruction 19-5
The director of the MBA program of a university wanted to know if a one week orientation would change the proportion among potential incoming students who would perceive the program as being good. Given below is the result from 215 students' view of the program before and after the orientation.
After the Orientation
 Before the Orientation  Good  Not Good  Total  Good 9337130 Not Good 711485 Total 16451215\begin{array} { | c | c | c | c | } \hline \text { Before the Orientation } & \text { Good } & \text { Not Good } & \text { Total } \\\hline \text { Good } & 93 & 37 & 130 \\\hline \text { Not Good } & 71 & 14 & 85 \\\hline \text { Total } & 164 & 51 & 215 \\\hline\end{array}

-Referring to Instruction 19-5,the null hypothesis for the Friedman rank test for the difference in the means should be rejected at a 0.01 level of significance.
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23
If the assumptions of the one-way ANOVA F test are not met,which of the following test could be used?

A) Friedman rank test
B) Marascuilo procedure
C) Kruskal-Wallis test
D) McNemar test
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24
The Kruskal-Wallis rank test for differences in more than two medians is a nonparametric alternative to ____

A) Student's t test for independent samples
B) Wilcoxon's rank sum test for differences in two medians
C) ANOVA F test for completely randomised experiments
D) Student's t test for related samples
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25
Instruction 19-5
The director of the MBA program of a university wanted to know if a one week orientation would change the proportion among potential incoming students who would perceive the program as being good. Given below is the result from 215 students' view of the program before and after the orientation.
After the Orientation
 Before the Orientation  Good  Not Good  Total  Good 9337130 Not Good 711485 Total 16451215\begin{array} { | c | c | c | c | } \hline \text { Before the Orientation } & \text { Good } & \text { Not Good } & \text { Total } \\\hline \text { Good } & 93 & 37 & 130 \\\hline \text { Not Good } & 71 & 14 & 85 \\\hline \text { Total } & 164 & 51 & 215 \\\hline\end{array}

-Referring to Instruction 19-5,the decision made at a 0.01 level of significance on the Friedman rank test for the difference in medians implies that the three medians are not all the same.
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26
Instruction 19-5
The director of the MBA program of a university wanted to know if a one week orientation would change the proportion among potential incoming students who would perceive the program as being good. Given below is the result from 215 students' view of the program before and after the orientation.
After the Orientation
 Before the Orientation  Good  Not Good  Total  Good 9337130 Not Good 711485 Total 16451215\begin{array} { | c | c | c | c | } \hline \text { Before the Orientation } & \text { Good } & \text { Not Good } & \text { Total } \\\hline \text { Good } & 93 & 37 & 130 \\\hline \text { Not Good } & 71 & 14 & 85 \\\hline \text { Total } & 164 & 51 & 215 \\\hline\end{array}

-Referring to Instruction 19-5,the decision made at a 0.01 level of significance on the Friedman rank test for the difference in medians implies that all three medians are significantly different.
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27
Instruction 19-5
The director of the MBA program of a university wanted to know if a one week orientation would change the proportion among potential incoming students who would perceive the program as being good. Given below is the result from 215 students' view of the program before and after the orientation.
After the Orientation
 Before the Orientation  Good  Not Good  Total  Good 9337130 Not Good 711485 Total 16451215\begin{array} { | c | c | c | c | } \hline \text { Before the Orientation } & \text { Good } & \text { Not Good } & \text { Total } \\\hline \text { Good } & 93 & 37 & 130 \\\hline \text { Not Good } & 71 & 14 & 85 \\\hline \text { Total } & 164 & 51 & 215 \\\hline\end{array}

-Referring to Instruction 19-5,the Friedman rank test is valid only if there is no interaction between the five blocks and the three treatment levels.
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28
Instruction 19-3
A perfume manufacturer is trying to choose between two magazine advertising layouts. An expensive layout would include a small package of the perfume. A cheaper layout would include a 'scratch-and-sniff' sample of the product. The manufacturer would use the more expensive layout only if there is evidence that it would lead to a higher approval rate. The manufacturer presents both layouts to five groups and determines the approval rating from each group on both layouts. The data are given below. Use this to test whether the median difference in approval rating is different from zero in favour of the more expensive layout with a level of significance of 0.05.
 Padkage  Scratch 52376843435348395647\begin{array}{cc}\text { Padkage } & \text { Scratch } \\\hline 52 & 37 \\68 & 43 \\43 & 53 \\48 & 39 \\56 & 47\end{array}


-Referring to Instruction 19-3,what is the rank of the absolute difference for the last pair of observations?
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29
The Kruskal-Wallis test is an extension of which of the following for two independent samples?

A) Paired-sample t test
B) Wilcoxon rank sum test
C) Pooled-variance t test
D) Wilcoxon signed ranks test
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30
Instruction 19-3
A perfume manufacturer is trying to choose between two magazine advertising layouts. An expensive layout would include a small package of the perfume. A cheaper layout would include a 'scratch-and-sniff' sample of the product. The manufacturer would use the more expensive layout only if there is evidence that it would lead to a higher approval rate. The manufacturer presents both layouts to five groups and determines the approval rating from each group on both layouts. The data are given below. Use this to test whether the median difference in approval rating is different from zero in favour of the more expensive layout with a level of significance of 0.05.
 Padkage  Scratch 52376843435348395647\begin{array}{cc}\text { Padkage } & \text { Scratch } \\\hline 52 & 37 \\68 & 43 \\43 & 53 \\48 & 39 \\56 & 47\end{array}


-Referring to Instruction 19-3,what are the lower and upper critical values of the test?
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31
Instruction 19-5
The director of the MBA program of a university wanted to know if a one week orientation would change the proportion among potential incoming students who would perceive the program as being good. Given below is the result from 215 students' view of the program before and after the orientation.
After the Orientation
 Before the Orientation  Good  Not Good  Total  Good 9337130 Not Good 711485 Total 16451215\begin{array} { | c | c | c | c | } \hline \text { Before the Orientation } & \text { Good } & \text { Not Good } & \text { Total } \\\hline \text { Good } & 93 & 37 & 130 \\\hline \text { Not Good } & 71 & 14 & 85 \\\hline \text { Total } & 164 & 51 & 215 \\\hline\end{array}

-Referring to Instruction 19-5,the Friedman rank test is valid only if the five blocks are independent so that the yields in one block have no influence on the yields in any other block.
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32
The Journal of Business Venturing reported on the activities of entrepreneurs during the organisation creation process.As part of a designed study,a total of 71 entrepreneurs were interviewed and divided into three groups: those that were successful in founding a new firm (n1 = 34),those still actively trying to establish a firm (n2 = 21),and those who tried to start a new firm but eventually gave up (n3 = 16).The total number of activities undertaken (e.g.,developed a business plan,sought funding,looked for facilities)by each group over a specified time period during organisation creation was measured.The objective is to compare the mean number of activities of the three groups of entrepreneurs.Because of concerns over necessary assumption of the parametric analysis,it was decided to use a nonparametric analysis.Identify the nonparametric method that would be used to analyse the data.

A) Kruskal-Wallis rank test for differences in medians
B) Wilcoxon signed rank test
C) Wilcoxon rank sum test
D) One-way ANOVA F test
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33
Instruction 19-3
A perfume manufacturer is trying to choose between two magazine advertising layouts. An expensive layout would include a small package of the perfume. A cheaper layout would include a 'scratch-and-sniff' sample of the product. The manufacturer would use the more expensive layout only if there is evidence that it would lead to a higher approval rate. The manufacturer presents both layouts to five groups and determines the approval rating from each group on both layouts. The data are given below. Use this to test whether the median difference in approval rating is different from zero in favour of the more expensive layout with a level of significance of 0.05.
 Padkage  Scratch 52376843435348395647\begin{array}{cc}\text { Padkage } & \text { Scratch } \\\hline 52 & 37 \\68 & 43 \\43 & 53 \\48 & 39 \\56 & 47\end{array}


-Referring to Instruction 19-3,the hypotheses that should be used are

A) H0: MD = 0 versus H1: MD ≠\neq 0.
B) H0: M1 ≠\neq 0 versus H1: M1 ≠\neq M2.
C)H0: MD ≤\le 0 versus H1: MD > 0.
D)H0: M1 ≤\le M2 versus H1: M1 > M2.
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34
Instruction 19-3
A perfume manufacturer is trying to choose between two magazine advertising layouts. An expensive layout would include a small package of the perfume. A cheaper layout would include a 'scratch-and-sniff' sample of the product. The manufacturer would use the more expensive layout only if there is evidence that it would lead to a higher approval rate. The manufacturer presents both layouts to five groups and determines the approval rating from each group on both layouts. The data are given below. Use this to test whether the median difference in approval rating is different from zero in favour of the more expensive layout with a level of significance of 0.05.
 Padkage  Scratch 52376843435348395647\begin{array}{cc}\text { Padkage } & \text { Scratch } \\\hline 52 & 37 \\68 & 43 \\43 & 53 \\48 & 39 \\56 & 47\end{array}


-Referring to Instruction 19-3,what is the value of the test statistic?
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35
Instruction 19-4
An agronomist wants to compare the crop yield of three varieties of chickpea seeds. She plants all three varieties of the seeds on each of five different patches of fields. She then measures the crop yield in bushels per acre. Treating this as a randomised block design, the results are presented in the table that follows.
 Fields  Smith  Walsh  Trevor 111.119.014.6213.518.015.7315.319.816.8414.619.616.759.816.615.2\begin{array} { | c | c | c | c | } \hline \text { Fields } & \text { Smith } & \text { Walsh } & \text { Trevor } \\\hline 1 & 11.1 & 19.0 & 14.6 \\\hline 2 & 13.5 & 18.0 & 15.7 \\\hline 3 & 15.3 & 19.8 & 16.8 \\\hline 4 & 14.6 & 19.6 & 16.7 \\\hline 5 & 9.8 & 16.6 & 15.2 \\\hline\end{array} Below is the Minitab output of the Friedman rank test:
Friectman Test Yedd versus Varieties Felds
Friedmantest for Yield by Varieties blocked by Fields S=10.00DF=2P=0.007S = 10.00 D F = 2 P = 0.007

 Varieties N EstMedian  Sum of Ranks  Smith 513.5005.0 Trevor 515.66710.0 Walsh 518.53315.0\begin{array} { l l l l } \text { Varieties } & N & \text { EstMedian } & \text { Sum of Ranks } \\ & & & \\ \text { Smith } & 5 & 13.500 & 5.0 \\ \text { Trevor } & 5 & 15.667 & 10.0 \\ \text { Walsh } & 5 & 18.533 & 15.0 \end{array}

 Grand median =15.900\text { Grand median }=15.900


-Referring to Instruction 19-4,the null hypothesis for the Friedman rank test is

A) MSmith = MWalsh = MTrevor.
B)H0: M Field 1 = M Field 2 = MField 3 = MField 4 = MField 5.
C)H0: μ\mu Field 1 = μ\mu Field 2 = μ\mu Field 3 = μ\mu Field 4 = μ\mu Field 5.
D) H0: μ\mu Smith = μ\mu Walsh = μ\mu Trevor.
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36
Suppose there is interest in comparing the median response time for three independent groups learning a specific task.The appropriate nonparametric procedure is ____.

A) Wilcoxon signed rank test
B) Kruskal-Wallis rank test for differences in medians
C) Wilcoxon rank sum test
D) Friedman rank test
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37
If the sample sizes in each group are more than five,the Kruskal-Wallis rank test statistic can be approximated by a standardised normal distribution.
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38
If the sample sizes in each group are more than five,the Kruskal-Wallis rank test statistic can be approximated by a chi-square distribution.
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39
Instruction 19-3
A perfume manufacturer is trying to choose between two magazine advertising layouts. An expensive layout would include a small package of the perfume. A cheaper layout would include a 'scratch-and-sniff' sample of the product. The manufacturer would use the more expensive layout only if there is evidence that it would lead to a higher approval rate. The manufacturer presents both layouts to five groups and determines the approval rating from each group on both layouts. The data are given below. Use this to test whether the median difference in approval rating is different from zero in favour of the more expensive layout with a level of significance of 0.05.
 Padkage  Scratch 52376843435348395647\begin{array}{cc}\text { Padkage } & \text { Scratch } \\\hline 52 & 37 \\68 & 43 \\43 & 53 \\48 & 39 \\56 & 47\end{array}


-Referring to Instruction 19-3,which pair(s)of observations has a negative signed rank?
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40
Instruction 19-3
A perfume manufacturer is trying to choose between two magazine advertising layouts. An expensive layout would include a small package of the perfume. A cheaper layout would include a 'scratch-and-sniff' sample of the product. The manufacturer would use the more expensive layout only if there is evidence that it would lead to a higher approval rate. The manufacturer presents both layouts to five groups and determines the approval rating from each group on both layouts. The data are given below. Use this to test whether the median difference in approval rating is different from zero in favour of the more expensive layout with a level of significance of 0.05.
 Padkage  Scratch 52376843435348395647\begin{array}{cc}\text { Padkage } & \text { Scratch } \\\hline 52 & 37 \\68 & 43 \\43 & 53 \\48 & 39 \\56 & 47\end{array}


-Referring to Instruction 19-3,what is the appropriate test to use?

A) Wilcoxon signed rank test for difference in median
B) Wilcoxon signed rank test for median difference
C) Wilcoxon rank sum test for median difference
D) Wilcoxon rank sum test for difference in median
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41
If the number of blocks in the experiment is more than five,the Friedman rank test statistic can be approximated by a standardised normal distribution.
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42
A non-parametric test for testing the difference between two medians from independent samples is known as the ____.

A) Kruskal-Wallis test
B) Wilcoxon rank sum test
C) Friedman rank test
D) Wilcoxon signed ranks test
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43
The ____ finds whether multiple sample groups have been selected from populations with equal medians.

A) Kruskal-Wallis test
B) Wilcoxon rank sum test
C) Friedman rank test
D) Wilcoxon signed ranks test
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44
If the number of blocks in the experiment is more than five,the Friedman rank test statistic can be approximated by a chi-square distribution.
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45
Instruction 19-6
An agronomist wants to compare the crop yield of three varieties of chickpea seeds. She plants all three varieties of the seeds on each of five different patches of fields. She then measures the crop yield in bushels per acre. Treating this as a randomised block design, the results are presented in the table that follows.
 Fields  Smith  Walsh  Trevor 111.119.014.6213.518.015.7315.319.816.8414.619.616.759.816.615.2\begin{array} { | c | c | c | c | } \hline \text { Fields } & \text { Smith } & \text { Walsh } & \text { Trevor } \\\hline 1 & 11.1 & 19.0 & 14.6 \\\hline2 & 13.5 & 18.0 & 15.7 \\\hline 3 & 15.3 & 19.8 & 16.8 \\\hline 4 & 14.6 & 19.6 & 16.7 \\\hline 5& 9.8 & 16.6 & 15.2 \\\hline\end{array} Below is the Minitab output of the Friedman rank test:
Friectman Test Yedd versus Varieties Felds
Friedmantest for Yield by Varieties blocked by Fields S=10.00DF=2P=0.007S = 10.00 D F = 2 P = 0.007

 Varieties N EstMedian  Sum of Ranks  Smith 513.5005.0 Trevor 515.66710.0 Walsh 518.53315.0\begin{array} { l l l l } \text { Varieties } & N & \text { EstMedian } & \text { Sum of Ranks } \\ & & & \\ \text { Smith } & 5 & 13.500 & 5.0 \\ \text { Trevor } & 5 & 15.667 & 10.0 \\ \text { Walsh } & 5 & 18.533 & 15.0 \end{array}

 Grand median =15.900\text { Grand median }=15.900


-Referring to Instruction 19-6,what is the value of the test statistic for the Friedman rank test for the difference in the medians?
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46
Instruction 19-6
An agronomist wants to compare the crop yield of three varieties of chickpea seeds. She plants all three varieties of the seeds on each of five different patches of fields. She then measures the crop yield in bushels per acre. Treating this as a randomised block design, the results are presented in the table that follows.
 Fields  Smith  Walsh  Trevor 111.119.014.6213.518.015.7315.319.816.8414.619.616.759.816.615.2\begin{array} { | c | c | c | c | } \hline \text { Fields } & \text { Smith } & \text { Walsh } & \text { Trevor } \\\hline 1 & 11.1 & 19.0 & 14.6 \\\hline2 & 13.5 & 18.0 & 15.7 \\\hline 3 & 15.3 & 19.8 & 16.8 \\\hline 4 & 14.6 & 19.6 & 16.7 \\\hline 5& 9.8 & 16.6 & 15.2 \\\hline\end{array} Below is the Minitab output of the Friedman rank test:
Friectman Test Yedd versus Varieties Felds
Friedmantest for Yield by Varieties blocked by Fields S=10.00DF=2P=0.007S = 10.00 D F = 2 P = 0.007

 Varieties N EstMedian  Sum of Ranks  Smith 513.5005.0 Trevor 515.66710.0 Walsh 518.53315.0\begin{array} { l l l l } \text { Varieties } & N & \text { EstMedian } & \text { Sum of Ranks } \\ & & & \\ \text { Smith } & 5 & 13.500 & 5.0 \\ \text { Trevor } & 5 & 15.667 & 10.0 \\ \text { Walsh } & 5 & 18.533 & 15.0 \end{array}

 Grand median =15.900\text { Grand median }=15.900


-Referring to Instruction 19-6,what are the degrees of freedom of the Friedman rank test for the difference in the medians at a level of significance of 0.01?
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47
When the normality assumption is not met in a randomised block design,which of the following tests should be used?

A) Kruskal-Wallis test
B) Wilcoxon rank sum test
C) Friedman rank test
D) Wilcoxon signed ranks test
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48
A non-parametric test for testing the median difference for paired samples is known as the ____.

A) Kruskal-Wallis test
B) Wilcoxon rank sum test
C) Friedman rank test
D) Wilcoxon signed ranks test
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49
A non-parametric test for testing the difference between two proportions from related samples is known as the ____.

A) Kruskal-Wallis test
B) Wilcoxon rank sum test
C) McNemar test
D) Wilcoxon signed ranks test
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50
Instruction 19-6
An agronomist wants to compare the crop yield of three varieties of chickpea seeds. She plants all three varieties of the seeds on each of five different patches of fields. She then measures the crop yield in bushels per acre. Treating this as a randomised block design, the results are presented in the table that follows.
 Fields  Smith  Walsh  Trevor 111.119.014.6213.518.015.7315.319.816.8414.619.616.759.816.615.2\begin{array} { | c | c | c | c | } \hline \text { Fields } & \text { Smith } & \text { Walsh } & \text { Trevor } \\\hline 1 & 11.1 & 19.0 & 14.6 \\\hline2 & 13.5 & 18.0 & 15.7 \\\hline 3 & 15.3 & 19.8 & 16.8 \\\hline 4 & 14.6 & 19.6 & 16.7 \\\hline 5& 9.8 & 16.6 & 15.2 \\\hline\end{array} Below is the Minitab output of the Friedman rank test:
Friectman Test Yedd versus Varieties Felds
Friedmantest for Yield by Varieties blocked by Fields S=10.00DF=2P=0.007S = 10.00 D F = 2 P = 0.007

 Varieties N EstMedian  Sum of Ranks  Smith 513.5005.0 Trevor 515.66710.0 Walsh 518.53315.0\begin{array} { l l l l } \text { Varieties } & N & \text { EstMedian } & \text { Sum of Ranks } \\ & & & \\ \text { Smith } & 5 & 13.500 & 5.0 \\ \text { Trevor } & 5 & 15.667 & 10.0 \\ \text { Walsh } & 5 & 18.533 & 15.0 \end{array}

 Grand median =15.900\text { Grand median }=15.900


-Referring to Instruction 19-6,what is the critical value of the Friedman rank test for the difference in the medians at a level of significance of 0.01?
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