Deck 17: Additional Tests for Nominal Data: Chi-Squared Tests
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Deck 17: Additional Tests for Nominal Data: Chi-Squared Tests
1
Which of the following uses a contingency table? 

D
2
Which of the following would be used to analyse the relationship between two categorical variables? 

D
3
Which of the following best describes the degrees of freedom needed in a Chi-squared test for normality? 

C
4
Which of the following best describes a Chi-squared goodness-of-fit test? 

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5
Which of the following may be used for hypothesis tests of nominal (categorical) data? 

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6
If each element in a population is classified into one and only one of several categories, the population is a: 

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7
A chi-squared test of a contingency table with 4 rows and 4 columns shows that the value of the test statistic is 23. Which of the following best describes the p-value for this test? 

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8
Which of the following uses the Chi-square distribution? 

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9
Which of the following may be used to determine whether data were drawn a particular distribution? 

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10
Of the values for a chi-squared test statistic listed below, which one is likely to lead to rejection of the null hypothesis in a goodness-of-fit test? 

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11
Which statistical technique is appropriate when we describe a single population of nominal data with exactly two categories? 

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12
Which statistical technique is appropriate when we wish to analyse the relationship between two nominal variables with two or more categories? 

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13
In a goodness-of-fit test, suppose that a sample showed that the observed frequency
and expected frequency
were equal for each cell i. Which of the following best describes the decision for the hypothesis test? 



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14
Which of the following should be used if we want to conduct a two-tail test of a population proportion? 

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15
Which of the following is the correct decision in a goodness-of-fit test, if the calculated value of the test statistic is 20.08 and there are 10 categories, testing at α of 1%? 

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16
Which of the following best describes the approach taken if the expected frequency
for any cell i is less than 5? 


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17
The vice-chancellor of a university collected data from students concerning the building of a new library, and classified the responses into different categories (strongly agree, agree, undecided, disagree, strongly disagree) and according to whether the student was male or female. Which of the following tests should be used to test whether responses and gender are independent? 

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18
Which of the following best describes the number of degrees of freedom used in a Chi-squared test of independence if there is a contingency table with 5 rows and 3 columns? 

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19
To determine the critical values in the chi-squared distribution table, the process requires which of the following information? 

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20
Which of the following are the degrees of freedom used in a Chi-squared test of independence between gender and mode of transport to university, if the mode of transport choices are public transport, car, bicycle or other. 

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21
Which statistical technique is appropriate when we compare two or more populations of nominal data with two or more categories? 

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22
Which of the following statements is not correct?
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23
Which of the following is not a characteristic of a multinomial experiment?



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24
Which of the following best describes the sampling distribution of the test statistic for a goodness-of-fit test with k categories? 

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25
Which statistical technique is appropriate when we describe a single population of nominal data with two or more categories? 

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26
A chi-squared test for independence with 6 degrees of freedom results in a test statistic
. Using the
tables, the most accurate statement that can be made about the p-value for this test is that: 



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27
Which statistical technique is appropriate when we compare two populations of nominal data with exactly two categories? 

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28
Which of the following statements is not correct? 

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29
Which of the following tests do not use the Chi-squared distribution? 

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30
The chi-squared test of a contingency table is based upon: 

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31
The number of degrees of freedom in a test of a contingency table with 7 rows and 5 columns is: 

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32
In a goodness-of-fit test, the null hypothesis states that the data came from a normally distributed population. The researcher estimated the population mean and population standard deviation from a sample of 500 observations. In addition, the researcher used 6 standardised intervals to test for normality. Using a 5% level of significance, the critical value for this test is: 

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33
In chi-squared tests, the conventional and conservative rule - known as the rule of five - is to require that the: 

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34
Consider a multinomial experiment with 100 trials, and the outcome of each trial can be classified into one of 4 categories. The number of degrees of freedom associated with the chi-squared goodness-of-fit test is: 

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35
In a chi-squared test of a contingency table, the value of the test statistic was , the significance level was = 0.05 and the degrees of freedom was 6. Thus: A. we fail to reject the null hyp othesis at .
B. we reject the null hypothesis at .
C. we don't have enough evidence to accept or reject the null hypothesis at .
D. we should increase the level of significance in order to reject the null hypothesis.
B. we reject the null hypothesis at .
C. we don't have enough evidence to accept or reject the null hypothesis at .
D. we should increase the level of significance in order to reject the null hypothesis.
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36
In a chi-squared goodness-of-fit test, if the expected frequencies and the observed frequencies were quite different, which of the following is the best conclusion?
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37
Which of the following statements is true for the chi-squared tests?
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38
Which of the following best describes the rejection region in a chi-squared test of independence? 

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39
The degrees of freedom in a chi-squared test for normality, where the number of standardised intervals is 13 and there are 2 population parameters to be estimated from the data, is equal to: 

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40
A left-tailed area in the chi-squared distribution equals 0.975. For df = 11, the table value equals: 

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41
Whenever the expected frequency of a cell is less than 5, one remedy for this condition is to increase the size of the sample.
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42
For a chi-squared distributed random variable with 12 degrees of freedom and a level of significance of 0.05, the chi-squared value from the table is 21.0261. The computed value of the test statistics is 25.1687. This will lead us to reject the null hypothesis.
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43
The middle 0.95 portion of the chi-squared distribution with 9 degrees of freedom has table values of 3.32511 and 16.9190, respectively.
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44
A left-tailed area in the chi-squared distribution equals 0.10. For 5 degrees of freedom, the table value equals 9.23635.
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45
A left-tailed area in the chi-squared distribution equals 0.90. For 10 degrees of freedom, the table value equals 15.9871.
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46
In a chi-squared test of independence, the value of the test statistic was 15.652, and the critical value at was 11.1433. Thus we must reject the null hypothesis at .
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47
A chi-squared goodness-of-fit test can be conducted either as a two-tail test or as a one-tail test.
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48
Whenever the expected frequency of a cell is less than 5, one possible remedy for this condition is to combine it with one or more other cells.
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49
A right-tailed area in the chi-squared distribution equals 0.01. For 4 degrees of freedom, the table value equals 13.2767.
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50
Whenever the expected frequency of a cell is less than 5, we must increase the significance level.
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51
A test for independence is applied to a contingency table with 4 rows and 4 columns for two nominal variables. The number of degrees of freedom for this test will be 9.
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52
In a goodness-of-fit test, the null hypothesis states that the data came from a normally distributed population. The researcher estimated the population mean and population standard deviation from a sample of 100 observations. In addition, the researcher used 6 standardised intervals to test for normality. Using a 2.5% level of significance, the critical value for this test is 9.3484.
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53
A Chi-squared test can be used to test the equality of two population means.
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54
One type of Chi-squared goodness of fit test is a Chi-squared test for normality.
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55
A test for independence is applied to a contingency table with 5 rows and 2 columns for two nominal variables. The number of degrees of freedom for this chi-squared test must be 4.
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56
The chi-squared test of independence is a Chi-squared test of a contingency table.
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57
A chi-squared test for independence with 6 degrees of freedom results in a test statistic of 13.25. Using the chi-squared table, the most accurate statement that can be made about the p-value for this test is that 0.025 < p-value < 0.05.
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58
A chi-squared test for independence with 10 degrees of freedom results in a test statistic of 17.894. Using the chi-squared table, the most accurate statement that can be made about the p-value for this test is that 0.05 < p-value < 0.10.
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59
For a chi-squared distributed random variable with 10 degrees of freedom and a level of significance of 0.025, the chi-squared value from the table is 20.5. The computed value of the test statistic is 16.857. The decision is to retain Ho.
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60
The null hypothesis states that the sample data came from a normally distributed population. The researcher calculates the sample mean and the sample standard deviation from the data. The data arrangement consisted of seven categories. Using a 0.05 significance level, the appropriate critical value for this chi-squared test for normality is 11.0705.
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61
In chi-squared tests, the conventional and conservative rule - known as the rule of five - is to require that difference between the observed and expected frequency for each cell be at least 5.
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62
An Australian firm has been accused of engaging in prejudicial hiring practices against women who have young children, where a young child is defined to be a child less than 12 years of age. According to the most recent census, the percentages of men, women without young children and women with young children in a certain community are 72%, 10% and 18%, respectively. A random sample of 200 employees of the firm revealed that 165 were men, 14 were women without young children and 21 were women with young children. Do the data provide sufficient evidence to conclude at the 10% level of significance that the firm has been engaged in prejudicial hiring practices?
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63
If we want to test for differences between two populations of nominal data with exactly two categories, we can employ either the z-test of , or the chi-squared test of a contingency table. (Squaring the value of the z-statistic yields the value of the -statistic.)
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64
The area to the right of a chi-squared value is 0.01. For 8 degrees of freedom, the table value is 20.0902.
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65
If we want to perform a one-tail test for differences between two populations of nominal data with exactly two categories, we must employ the z-test of .
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66
The rule of five is used to ensure that the discrete distribution of the test statistic can be approximated by the continuous Chi-squared distribution.
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67
The number of degrees of freedom associated with the chi-squared test for normality is the number of intervals used minus the number of parameters estimated from the data.
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68
At present in Australia, voting in a Federal election is compulsory for all citizens 18 years or over.
There has been some discussion to have this changed from being compulsory to voluntary.
A study was done to investigate whether there was a relationship between gender and attitude towards changing voting in Australia from compulsory to voluntary. Participants were asked "Do you agree to change voting in Australian Federal Elections from compulsory to voluntary? A table of results from the sample is given below.
At the 1% level of significance, is there a relationship between gender and attitude towards voluntary voting?
There has been some discussion to have this changed from being compulsory to voluntary.
A study was done to investigate whether there was a relationship between gender and attitude towards changing voting in Australia from compulsory to voluntary. Participants were asked "Do you agree to change voting in Australian Federal Elections from compulsory to voluntary? A table of results from the sample is given below.
At the 1% level of significance, is there a relationship between gender and attitude towards voluntary voting?

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69
The personnel manager of a consumer product company asked a random sample of employees how they felt about the work they were doing. The following table gives a breakdown of their responses by gender.
Do the data provide sufficient evidence to conclude that the level of job satisfaction is related to gender? Use
0.10.


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70
A sport preference poll showed the following data for men and women:
Using the 5% level of significance, test to determine whether sport preferences depend on gender.

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71
The following data are believed to have come from a normal probability distribution.
The mean of this sample equals 26.80, and the standard deviation equals 6.378. Use the goodness-of-fit test at the 5% significance level to test this claim.

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72
Like that of the Student t-distribution, the shape of the chi-squared distribution depends on its number of degrees of freedom.
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73
The number of degrees of freedom for a contingency table with r rows and c columns is (r -1)(c - 1), provided that both r and c follow the rule of five.
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74
When doing a Chi-squared goodness of fit test, we use Ho: The data are not normally distributed.
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75
If we want to perform a two-tail test for differences between two populations of nominal data with exactly two categories, we can employ either the z-test of , or the chi-squared test of a contingency table. (Squaring the value of the z-statistic yields the value of the -statistic.)
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76
In a goodness-of-fit test, the null hypothesis states that the data came from a normally distributed population. The researcher estimated the population mean and population standard deviation from a sample of 200 observations. In addition, the researcher used 5 standardised intervals to test for normality. Using a 10% level of significance, the critical value for this test is 4.60517.
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77
The chi-squared goodness-of-fit test is usually used as a test of multinomial parameters, but it can also be used to determine whether data were drawn from any distribution.
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78
The chi-squared test of a contingency table is used to determine if there is enough evidence to infer that two nominal variables are related, and to infer that differences exist among two or more populations of nominal variables.
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79
A multinomial experiment, where the outcome of each trial can be classified into one of two categories, is identical to the binomial experiment.
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80
Five brands of orange juice are displayed side by side in several supermarkets in a large city. It was noted that in one day, 180 customers purchased orange juice. Of these, 30 picked Brand A, 40 picked Brand B, 25 picked Brand C, 35 picked Brand D, and 50 picked brand
E. In this city, can you conclude at the 5% significance level that there is a preferred brand of orange juice?
E. In this city, can you conclude at the 5% significance level that there is a preferred brand of orange juice?
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