Deck 15: Nonparametric Methods: Nominal Level Hypothesis Tests
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Deck 15: Nonparametric Methods: Nominal Level Hypothesis Tests
1
A question has these possible responses: excellent,very good,good,fair,and unsatisfactory. What are the degrees of freedom for a goodness-of-fit test to test the hypothesis that responses are uniformly distributed?
A)0
B)2
C)4
D)5
A)0
B)2
C)4
D)5
4
2
For a contingency table,the expected frequency for a cell is found by dividing the row total by the grand total.
False
3
A t-statistic is useful for computing an expected normal distribution.
False
4
There is not one,but a family of chi-square distributions. There is a chi-square distribution for 1 degree of freedom,another for 2 degrees of freedom,another for 3 degrees of freedom,and so on.
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5
To test the null hypothesis that a set of sample data is normally distributed,we compare an expected normal distribution of the data to an observed distribution of the data.
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6
The shape of the chi-square distribution depends on the size of the sample.
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7
If we are testing the difference between two population proportions,it is assumed that the two populations are approximately normal and have equal variances.
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8
In the goodness-of-fit test,the chi-square distribution is used to determine how well an observed distribution of observations "fits" an expected distribution of observations.
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9
The pooled estimate of the proportion is found by dividing the total number of samples by the total number of successes.
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10
The shape of the chi-square distribution depends on the number of degrees of freedom.
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11
Nonparametric tests require no assumptions about the shape of the population distribution.
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12
In testing the difference between two population proportions,we pool the two sample proportions to estimate the population proportion.
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13
What is the critical value at the 0.05 level of significance for a goodness-of-fit test if there are six categories?
A)3.841
B)5.991
C)7.815
D)11.070
A)3.841
B)5.991
C)7.815
D)11.070
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14
The chi-square distribution is positively skewed.
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15
The chi-square test statistic used in a goodness-of-fit test has k − 1 degrees of freedom.
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16
For a goodness-of-fit test,the following are possible null and alternate hypotheses:
H0: Sales are uniformly distributed among the five locations.
H1: Sales are not uniformly distributed among the five locations.
H0: Sales are uniformly distributed among the five locations.
H1: Sales are not uniformly distributed among the five locations.
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17
The claim that "male and female students at Coastal Carolina University prefer different parking lots on campus" is an example of a chi-square null hypothesis.
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18
Some important uses of the chi-square include testing for the association of two categorical variables.
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19
The variance of the chi-square distribution is equal to one.
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20
The use of the chi-square statistic would be permissible in the following goodness-of-fit test. 

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21
The chi-square distribution is ________.
A)positively skewed
B)negatively skewed
C)normally distributed
D)uniformly distributed
A)positively skewed
B)negatively skewed
C)normally distributed
D)uniformly distributed
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22
The contingency table for a sample of corporate executives classified by educational level and social activity follows.
What does the expected frequency for the "Above Average" social activity and "High School" education equal?
A)9.50
B)60.00
C)22.50
D)28.50

A)9.50
B)60.00
C)22.50
D)28.50
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23
A distributor of personal computers has five locations in a city. In the year's first quarter,the sales in units were:
For a goodness-of-fit test that sales were the same for all locations,what is the critical value at the 0.01 level of risk?
A)7.779
B)15.033
C)13.277
D)5.412

A)7.779
B)15.033
C)13.277
D)5.412
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24
A sample of 100 production workers is obtained. The workers are classified by gender (male,female)and by age (under 20,20−29,30−39,and 40 or over). How many degrees of freedom are there?
A)0
B)3
C)6
D)5
A)0
B)3
C)6
D)5
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25
What is the decision regarding the differences between the observed and expected frequencies if the critical value of the chi-square is 9.488 and the computed chi-square value is 6.079?
A)Fail to reject the null hypothesis; the difference is probably due to sampling error.
B)Reject the null hypothesis.
C)Fail to reject the alternate hypothesis.
D)It is too close; reserve judgment.
A)Fail to reject the null hypothesis; the difference is probably due to sampling error.
B)Reject the null hypothesis.
C)Fail to reject the alternate hypothesis.
D)It is too close; reserve judgment.
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26
Which of the following can be used to test the hypothesis that two nominal variables are related?
A)A contingency table analysis
B)A goodness-of-fit
C)ANOVA table
D)A regression analysis
A)A contingency table analysis
B)A goodness-of-fit
C)ANOVA table
D)A regression analysis
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27
The following table classifies 100 individuals using two variables,gender and college attended.
What is this two-way classification called?
A)Goodness-of-fit test
B)Frequency table
C)ANOVA table
D)Contingency table

A)Goodness-of-fit test
B)Frequency table
C)ANOVA table
D)Contingency table
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28
Three new colors have been proposed for the Jeep Grand Cherokee vehicle. They are silver blue,almond,and willow green. The null hypothesis for a goodness-of-fit test would be ________.
A)willow green is preferred over the other colors
B)there is no preference between the colors
C)any one color is preferred over the other colors
D)it is impossible to determine
A)willow green is preferred over the other colors
B)there is no preference between the colors
C)any one color is preferred over the other colors
D)it is impossible to determine
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29
Which of the following assumptions is necessary to apply a goodness-of-fit test?
A)The population must be normally distributed.
B)The data are measured with a nominal or ordinal scale.
C)The population variance must be known.
D)The population mean must be known.
A)The population must be normally distributed.
B)The data are measured with a nominal or ordinal scale.
C)The population variance must be known.
D)The population mean must be known.
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30
What are the degrees of freedom for a contingency table analysis?
A)n - 1
B)Rows − Columns
C)(Rows)× (Columns)
D)(Rows − 1)× (Columns − 1)
A)n - 1
B)Rows − Columns
C)(Rows)× (Columns)
D)(Rows − 1)× (Columns − 1)
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31
For people released from prison,the following table shows their adjustment to civilian life and place of residence.
What is the critical value for this contingency table at the 0.01 level of significance?
A)9.488
B)2.070
C)11.345
D)13.277

A)9.488
B)2.070
C)11.345
D)13.277
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32
The chi-square statistic has ________.
A)one distribution
B)two distributions
C)a family of distributions
D)a uniform distribution
A)one distribution
B)two distributions
C)a family of distributions
D)a uniform distribution
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33
The chi-square statistic ________.
A)is greater than or equal to zero
B)is less than or equal to zero
C)can be any value
D)is equal to zero
A)is greater than or equal to zero
B)is less than or equal to zero
C)can be any value
D)is equal to zero
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34
Which of the following statements is correct regarding the goodness-of-fit test?
A)Variables are based on the nominal measurement scale.
B)Population must be normal.
C)All the expected frequencies must be equal.
D)All the expected frequencies must be unequal.
A)Variables are based on the nominal measurement scale.
B)Population must be normal.
C)All the expected frequencies must be equal.
D)All the expected frequencies must be unequal.
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35
The chi-square distribution becomes more symmetrical as the ________.
A)number of variables increases
B)chi-square value increases
C)degrees of freedom decrease
D)degrees of freedom increase
A)number of variables increases
B)chi-square value increases
C)degrees of freedom decrease
D)degrees of freedom increase
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36
For a chi-square test involving a contingency table,suppose the null hypothesis is rejected. We conclude that the two variables are ________.
A)linear
B)curvilinear
C)not related
D)related
A)linear
B)curvilinear
C)not related
D)related
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37
What is our decision for a goodness-of-fit test with a computed chi-square value of 1.273 and a critical value of 13.388?
A)Do not reject the null hypothesis.
B)Reject the null hypothesis.
C)We are unable to reject or not reject the null hypothesis based on data.
D)We should take a larger sample.
A)Do not reject the null hypothesis.
B)Reject the null hypothesis.
C)We are unable to reject or not reject the null hypothesis based on data.
D)We should take a larger sample.
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38
For any chi-square goodness-of-fit test,the number of degrees of freedom is found by ________.
A)n - k − 1
B)k − 1
C)n + 1
D)n + k
A)n - k − 1
B)k − 1
C)n + 1
D)n + k
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39
Which of the following statements is correct regarding the chi-square distribution?
A)The distribution is negatively skewed.
B)Chi-square is based on two sets of degrees of freedom,one for the numerator and one for the denominator.
C)The shape of the distribution is based on the degrees of freedom.
D)The variance of the distribution is equal to one.
A)The distribution is negatively skewed.
B)Chi-square is based on two sets of degrees of freedom,one for the numerator and one for the denominator.
C)The shape of the distribution is based on the degrees of freedom.
D)The variance of the distribution is equal to one.
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40
Which chi-square distribution would be closest to a normal distribution?
A)The distribution with 3 degrees of freedom
B)The distribution with 12 degrees of freedom
C)The distribution with 15 degrees of freedom
D)The distribution with 9 degrees of freedom
A)The distribution with 3 degrees of freedom
B)The distribution with 12 degrees of freedom
C)The distribution with 15 degrees of freedom
D)The distribution with 9 degrees of freedom
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41
A recent study of the relationship between social activity and education for a sample of corporate executives showed the following results.
The degrees of freedom for the analysis are ________.
A)1
B)2
C)3
D)4

A)1
B)2
C)3
D)4
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42
A recent study of the relationship between social activity and education for a sample of corporate executives showed the following results.
The appropriate test statistic for the analysis is a(n)________.
A)F-statistic
B)t-statistic
C)chi-square statistic
D)z-statistic

A)F-statistic
B)t-statistic
C)chi-square statistic
D)z-statistic
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43
A recent study of the relationship between social activity and education for a sample of corporate executives showed the following results.
The null hypothesis for the analysis is ________.
A)there is no relationship between social activity and education
B)the correlation between social activity and education is zero
C)as social activity increases,education increases
D)the mean of social activity equals the mean of education

A)there is no relationship between social activity and education
B)the correlation between social activity and education is zero
C)as social activity increases,education increases
D)the mean of social activity equals the mean of education
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44
A recent study of the relationship between social activity and education for a sample of corporate executives showed the following results.
Based on the analysis,what can be concluded?
A)Social activity and education are correlated.
B)Social activity and education are not related.
C)Social activity and education are related.
D)No conclusion is possible.

A)Social activity and education are correlated.
B)Social activity and education are not related.
C)Social activity and education are related.
D)No conclusion is possible.
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45
Recently,students in a marketing research class were interested in the driving behavior of students. Specifically,the marketing students were interested in finding out if exceeding the speed limit was related to social activity. They collected the following responses from 100 randomly selected students:
Using 0.05 as the significance level,what is the critical value for the test statistic?
A)9.488
B)5.991
C)7.815
D)3.841

A)9.488
B)5.991
C)7.815
D)3.841
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46
A personnel manager is concerned about absenteeism. She decides to sample employee records to determine if absenteeism is distributed evenly throughout the six-day workweek. The null hypothesis is: Absenteeism is distributed evenly throughout the week. The 0.01 level is to be used. The sample results are:
What is the calculated value of chi-square?
A)1.0
B)0.5
C)0.8
D)8.0

A)1.0
B)0.5
C)0.8
D)8.0
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47
Recently,students in a marketing research class were interested in the driving behavior of students. Specifically,the marketing students were interested in finding out if exceeding the speed limit was related to social activity. They collected the following responses from 100 randomly selected students:
What is the value of the test statistic?
A)83.67
B)9.890
C)50
D)4.94

A)83.67
B)9.890
C)50
D)4.94
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48
Recently,students in a marketing research class were interested in the driving behavior of students. Specifically,the marketing students were interested in finding out if exceeding the speed limit was related to social activity. They collected the following responses from 100 randomly selected students:
Based on the analysis,what can be concluded?
A)Driving behavior and gender are correlated.
B)Driving behavior and gender are not related.
C)Driving behavior and gender are related.
D)No conclusion is possible.

A)Driving behavior and gender are correlated.
B)Driving behavior and gender are not related.
C)Driving behavior and gender are related.
D)No conclusion is possible.
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49
A recent study of the relationship between social activity and education for a sample of corporate executives showed the following results.
The appropriate test to analyze the relationship between social activity and education is ________.
A)a regression analysis
B)an analysis of variance
C)a contingency table analysis
D)a goodness-of-fit test

A)a regression analysis
B)an analysis of variance
C)a contingency table analysis
D)a goodness-of-fit test
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50
A personnel manager is concerned about absenteeism. She decides to sample employee records to determine if absenteeism is distributed evenly throughout the six-day workweek. The null hypothesis is: Absenteeism is distributed evenly throughout the week. The 0.01 level is to be used. The sample results are:
What is the expected frequency?
A)9
B)10
C)11
D)12

A)9
B)10
C)11
D)12
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51
A personnel manager is concerned about absenteeism. She decides to sample employee records to determine if absenteeism is distributed evenly throughout the six-day workweek. The null hypothesis is: Absenteeism is distributed evenly throughout the week. Use the 0.01 level of significance. The sample results are::
How many degrees of freedom are there?
A)0
B)3
C)4
D)5

A)0
B)3
C)4
D)5
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52
When determining how well an observed set of frequencies fits an expected set of frequencies,what is the test statistic?
A)F-statistic
B)t-statistic
C)X2-statistic
D)z-statistic
A)F-statistic
B)t-statistic
C)X2-statistic
D)z-statistic
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53
Recently,students in a marketing research class were interested in the driving behavior of students. Specifically,the marketing students were interested in finding out if exceeding the speed limit was related to gender. They collected the following responses from 100 randomly selected students:
The null hypothesis for the analysis is ________.
A)there is no relationship between gender and speeding
B)the correlation between driving behavior and gender is zero
C)as driving behavior increases,gender increases
D)the mean of driving behavior equals the mean of gender

A)there is no relationship between gender and speeding
B)the correlation between driving behavior and gender is zero
C)as driving behavior increases,gender increases
D)the mean of driving behavior equals the mean of gender
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54
Recently,students in a marketing research class were interested in the driving behavior of students. Specifically,the marketing students were interested in finding out if exceeding the speed limit was related to social activity. They collected the following responses from 100 randomly selected students:
The degrees of freedom for the analysis are ________.
A)1
B)2
C)3
D)4

A)1
B)2
C)3
D)4
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55
A personnel manager is concerned about absenteeism. She decides to sample employee records to determine if absenteeism is distributed evenly throughout the six-day workweek. The null hypothesis is: Absenteeism is distributed evenly throughout the week. The 0.01 level is to be used. The sample results are:
What is the critical value of a chi-square with α = 0.05?
A)11.070
B)12.592
C)13.388
D)15.033

A)11.070
B)12.592
C)13.388
D)15.033
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56
In a goodness-of-fit test,the null hypothesis (no difference between sets of observed and expected frequencies)is rejected when the ________.
A)computed chi-square is less than the critical value
B)difference between the observed and expected frequencies is significantly large
C)difference between the observed and expected frequencies is small
D)difference between the observed and expected frequencies occurs by chance
A)computed chi-square is less than the critical value
B)difference between the observed and expected frequencies is significantly large
C)difference between the observed and expected frequencies is small
D)difference between the observed and expected frequencies occurs by chance
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57
A recent study of the relationship between social activity and education for a sample of corporate executives showed the following results.
What is the value of the chi-square test statistic?
A)100
B)83.67
C)50
D)4.94

A)100
B)83.67
C)50
D)4.94
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58
A recent study of the relationship between social activity and education for a sample of corporate executives showed the following results.
Using 0.05 as the significance level,what is the critical value for the test statistic?
A)9.488
B)5.991
C)7.815
D)3.841

A)9.488
B)5.991
C)7.815
D)3.841
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59
A personnel manager is concerned about absenteeism. She decides to sample employee records to determine if absenteeism is distributed evenly throughout the six-day workweek. The null hypothesis is: Absenteeism is distributed evenly throughout the week. The 0.01 level is to be used. The sample results are:
What kind of frequencies are the numbers 12,9,11,10,9,and 9 called?
A)Acceptance frequencies
B)Critical frequencies
C)Expected frequencies
D)Observed frequencies

A)Acceptance frequencies
B)Critical frequencies
C)Expected frequencies
D)Observed frequencies
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60
The computed chi-square value is positive because the difference between the observed and expected frequencies is ________.
A)squared
B)linear
C)uniform
D)always positive
A)squared
B)linear
C)uniform
D)always positive
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61
To test if an observed frequency distribution with five classes is normally distributed,we need to find ________.
A)the t-statistic
B)the expected,normally distributed class frequencies
C)the class marks
D)the class relative frequencies
A)the t-statistic
B)the expected,normally distributed class frequencies
C)the class marks
D)the class relative frequencies
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62
A frequency distribution has a mean of 100 and a standard deviation of 20. The class limits for one class are 50 up to 60. Based on the normal distribution,what is the probability that an observation would be in this class?
A)0.4938
B)0.4772
C)0.0166
D)−0.0166
A)0.4938
B)0.4772
C)0.0166
D)−0.0166
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63
Suppose we test H0: π1 = π2 at the 0.05 level of significance. If the z-test statistic is −1.07,what is our decision?
A)Reject the null hypothesis.
B)Do not reject the null hypothesis.
C)Take a larger sample.
D)Reserve judgment.
A)Reject the null hypothesis.
B)Do not reject the null hypothesis.
C)Take a larger sample.
D)Reserve judgment.
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64
A frequency distribution has a mean of 100 and a standard deviation of 20. The class limits for one class are 50 up to 60. What are the standard normal z-statistics for the class limits?
A)−20 and 20
B)−2.5 and −2.0
C)2.0 and 2.5
D)−50 and −40
A)−20 and 20
B)−2.5 and −2.0
C)2.0 and 2.5
D)−50 and −40
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65
A survey of property owners' opinions about a street-widening project was taken to determine if owners' opinions were related to the distance between their home and the street. A randomly selected sample of 100 property owners was contacted and the results are shown next.
What is the expected frequency for people who are in favor of the project and have less than 45 feet of property foot frontage?
A)10
B)12
C)35
D)50

A)10
B)12
C)35
D)50
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66
A sample of 250 adults tried the new multigrain cereal Wow! A total of 187 rated it as excellent. In a sample of 100 children,66 rated it as excellent. Using the 0.1 significance level,the researcher wishes to show that adults like the cereal better than children. What is the pooled proportion?
A)0.723
B)1.408
C)0.494
D)0.807
A)0.723
B)1.408
C)0.494
D)0.807
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67
A survey of property owners' opinions about a street-widening project was taken to determine if owners' opinions were related to the distance between their home and the street. A randomly selected sample of 100 property owners was contacted and the results are shown next.
What is the critical value at the 10% level of significance?
A)7.779
B)9.236
C)9.488
D)11.070

A)7.779
B)9.236
C)9.488
D)11.070
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68
A survey of property owners' opinions about a street-widening project was taken to determine if owners' opinions were related to the distance between their home and the street. A randomly selected sample of 100 property owners was contacted and the results are shown next.
What is the critical value at the 5% level of significance?
A)7.779
B)9.488
C)9.236
D)11.070

A)7.779
B)9.488
C)9.236
D)11.070
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69
A frequency distribution has a mean of 200 and a standard deviation of 20. The class limits for one class are 220 up to 240. Based on the normal distribution,what is the probability that an observation would be in this class?
A)0.1359
B)0.3413
C)0.4772
D)0.8185
A)0.1359
B)0.3413
C)0.4772
D)0.8185
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70
To test if an observed frequency distribution with five classes is normally distributed,we compute probabilities for each class based on a(n)________.
A)standard normal distribution
B)chi-square distribution
C)student's t-distribution
D)F-distribution
A)standard normal distribution
B)chi-square distribution
C)student's t-distribution
D)F-distribution
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71
A survey of property owners' opinions about a street-widening project was taken to determine if owners' opinions were related to the distance between their home and the street. A randomly selected sample of 100 property owners was contacted and the results are shown next.
How many degrees of freedom are there?
A)2
B)3
C)4
D)5

A)2
B)3
C)4
D)5
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72
A survey of property owners' opinions about a street-widening project was taken to determine if owners' opinions were related to the distance between their home and the street. A randomly selected sample of 100 property owners was contacted and the results are shown next.
What is the expected frequency for people who are undecided about the project and have property front footage between 45 and 120 feet?
A)2.2
B)3.9
C)5.0
D)7.7

A)2.2
B)3.9
C)5.0
D)7.7
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73
How is a pooled estimate of the population proportion represented?
A)pc
B)z
C)π
D)nπ
A)pc
B)z
C)π
D)nπ
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74
A sample of 250 adults tried the new multigrain cereal Wow! A total of 187 rated it as excellent. In a sample of 100 children,66 rated it as excellent. Using the 0.1 significance level,the researcher wishes to show that adults like the cereal better than children. Which of the following is the alternate hypothesis?
A)H1: πA = πC
B)H1: πA < πC
C)H1: πA > πC
D)H1: πA ≠ πC
A)H1: πA = πC
B)H1: πA < πC
C)H1: πA > πC
D)H1: πA ≠ πC
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75
A frequency distribution has a mean of 200 and a standard deviation of 20. The class limits for one class are 220 up to 240. What are the standard normal z-statistics for the class limits?
A)−20 and 20
B)−2.0 and −1.0
C)200 and 220
D)1.0 and 2.0
A)−20 and 20
B)−2.0 and −1.0
C)200 and 220
D)1.0 and 2.0
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76
A survey of property owners' opinions about a street-widening project was taken to determine if owners' opinions were related to the distance between their home and the street. A randomly selected sample of 100 property owners was contacted and the results are shown next.
What is the expected frequency for people against the project and who have over 120 feet of property foot frontage?
A)1.1
B)3.9
C)5.0
D)5.5

A)1.1
B)3.9
C)5.0
D)5.5
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77
To test if an observed frequency distribution with five classes is normally distributed,we need to ________.
A)compute an F-statistic
B)calculate a t-statistic
C)convert the class marks to standard normal z-statistics
D)convert the class limits to standard normal z-statistics
A)compute an F-statistic
B)calculate a t-statistic
C)convert the class marks to standard normal z-statistics
D)convert the class limits to standard normal z-statistics
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78
In a market test of a new chocolate raspberry coffee,a poll of 400 people from Dobbs Ferry showed 250 preferred the new coffee. In Irvington,170 out of 350 people preferred the new coffee. To test the hypothesis that there is no difference in preferences between the two villages,what is the null hypothesis?
A)H0: π1 < π2
B)H0: π1 > π2
C)H0: π1 = π2
D)H0: π1 ≠ π2
A)H0: π1 < π2
B)H0: π1 > π2
C)H0: π1 = π2
D)H0: π1 ≠ π2
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79
If the decision is to reject the null hypothesis of no difference between two population proportions at the 5% level of significance,what are the alternative hypothesis and rejection region?
A)H1: π1 ≠ π2; z > +1.645 and z < −1.645
B)H1: π1 ≠ π2; z > +1.960 and z < −1.960
C)H1: π1 > π2; z < −1.645
D)H1: π1 > π2; z < −1.960
A)H1: π1 ≠ π2; z > +1.645 and z < −1.645
B)H1: π1 ≠ π2; z > +1.960 and z < −1.960
C)H1: π1 > π2; z < −1.645
D)H1: π1 > π2; z < −1.960
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80
In a market test of a new chocolate raspberry coffee,a poll of 400 people (sample 1)from Dobbs Ferry showed 250 preferred the new coffee. In Irvington,170 out of 350 people (sample 2)preferred the new coffee. To test the hypothesis that a higher proportion of people in Dobbs Ferry prefer the new coffee,what is the alternate hypothesis?
A)H1: π1 < π2
B)H1: π1 > π2
C)H1: π1 = π2
D)H1: π1 ≠ π2
A)H1: π1 < π2
B)H1: π1 > π2
C)H1: π1 = π2
D)H1: π1 ≠ π2
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