Deck 15: Nonparametric Methods: Nominal Level Hypothesis Tests
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Deck 15: Nonparametric Methods: Nominal Level Hypothesis Tests
1
In testing the difference between two population proportions, we pool the two sample proportions to estimate the population proportion.
True
2
The variance of the chi-square distribution is equal to one.
False
3
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.
True
4
The chi-square distribution is positively skewed.
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5
Nonparametric tests require no assumptions about the shape of the population distribution.
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6
The shape of the chi-square distribution depends on the number of degrees of freedom.
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7
The chi-square test statistic used in a goodness-of-fit test has k - 1 degrees of freedom.
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8
The computed value of a chi-square statistic is always positive because the numerator is the difference between the observed frequencies and the expected frequencies ______________.
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9
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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10
The use of the chi-square statistic would be permissible in the following goodness-of-fit test. 

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11
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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12
Some important uses of the chi-square include testing for the association of two categorical variables.
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13
For a contingency table, the expected frequency for a cell is found by dividing the row total by the grand total.
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14
The shape of the chi-square distribution depends on the size of the sample.
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15
A t-statistic is useful for computing an expected normal distribution.
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16
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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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
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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19
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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20
A chi-square goodness-of-fit test is used to determine how well an observed distribution fits an _________ distribution.
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21
In a contingency table, multiplying the row total by the column total and dividing by the ___________ computes the expected frequency for a cell.
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22
When testing a goodness-of-fit null hypothesis, and there are extremely large differences between observed and expected frequencies, the null hypothesis should be ___________.
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23
If the computed value of a chi-square statistic is greater than the critical value, the null hypothesis is ________.
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24
The purpose of pooling the sample proportions when testing the difference between two population proportions is to ___________.
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25
To verify that an observed frequency distribution is normally distributed, a ________ statistic is used to test the hypothesis that the sample data is normally distributed.
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26
The lowest level of measurement for a chi-square goodness-of-fit test is the ______________ level.
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27
The degrees of freedom for a chi-square goodness-of-fit test is equal to ______________.
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28
A contingency table shows the frequencies for three levels of income with gender. There are ______ degrees of freedom to test the null hypothesis that income and gender are independent.
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29
The shape of the chi-square distribution is _________________.
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30
The test statistic for a goodness-of-fit test for unequal expected frequencies is a __________ statistic.
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31
For any goodness-of-fit test, the null hypothesis is that there is ____ difference between the expected and observed distributions.
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32
The null hypothesis in the goodness-of-fit test is __________.
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33
To verify that a frequency distribution for sample data is normally distributed, the expected frequencies are computed using probabilities from a _____________ distribution.
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34
To test a hypothesis that a frequency distribution for sample data is normally distributed, class limits are transformed using a ____________.
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35
Contingency table analysis can be used to test for a relationship between two _________ scale variables.
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36
To test a hypothesis to verify that an observed frequency distribution data is normally distributed, the expected frequencies are _______________ distributed.
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37
As the degrees of freedom increase, the shape of a chi-square distribution approaches a ______________ distribution.
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38
In a contingency table analysis, the decision to reject the null hypothesis is based on a __________ test statistic.
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39
For hypothesis tests using a chi-square statistic, the rejection region is in the _____________ tail of the chi-square distribution.
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40
In a contingency table, the decision to reject the null hypothesis is based on a __________ test statistic.
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41
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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42
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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43
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) 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) Unable to reject or not reject the null hypothesis based on data.
D) We should take a larger sample.
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44
When testing the null hypothesis that two population proportions are equal, the hypothesized difference between the population proportions is ________________.
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45
If we are testing for the difference between two population proportions, it is assumed that the two populations follow the ________________ distribution.
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46
What is the decision regarding the differences between the observed and expected frequencies if the critical value of 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) 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) Too close; reserve judgment.
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47
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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48
Among 200 people surveyed, 66 people or 0.33% preferred the product. 0.33% is called the _______________.
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49
A manufacturer claims that less than 1% of all its products do not meet the minimum government standards. A survey of 500 products revealed that 10 did not meet the standard. If the z-statistic is -2.58 and the level of significance is 0.02, your decision would be __________.
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50
A manufacturer claims that less than 1% of all its products do not meet the minimum government standards. A survey of 500 products revealed that 10 did not meet the standard. If the computed z-statistic is -1.960 and the level of significance is 0.01, your decision would be _______________.
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51
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
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52
A manufacturer claims that less than 1% of all its products do not meet the minimum government standards. A survey of 500 products revealed that 10 did not meet the standard. If the level of significance is 2%, the critical value is _____.
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53
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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54
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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55
If we are testing for the difference between two population proportions, it is assumed that the two samples are large enough that the binomial distribution can be approximated by ______.
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56
A manufacturer claims that less than 1% of all its products do not meet the minimum government standards. A survey of 500 products revealed that 10 did not meet the standard. Thus, if α = .01, the critical value is ____________.
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57
A manufacturer claims that less than 1% of all its products do not meet the minimum government standards. A survey of 500 products revealed that 10 did not meet the standard. Thus, the computed z statistic is ______________.
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58
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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59
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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60
The pooled estimate of the proportion is found by dividing the total number of successes by ___________________________.
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61
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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62
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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63
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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64
The contingency table for a sample of corporate executives classified by educational level and the 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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65
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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66
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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67
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
D) A regression analysis
A) A contingency table analysis
B) A goodness-of-fit
C) ANOVA
D) A regression analysis
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68
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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69
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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70
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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71
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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72
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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73
The chi-square distribution becomes more symmetrical as the ____________.
A) Number of variables increases
B) The chi-square value increases
C) Degrees of freedom decrease
D) Degrees of freedom increase
A) Number of variables increases
B) The chi-square value increases
C) Degrees of freedom decrease
D) Degrees of freedom increase
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74
Three new colors have been proposed for the Jeep Grand Cherokee vehicle. They are silvered-blue, almond, and willow green. The null hypothesis for a goodness-of-fit test would be _______________.
A) That willow green is preferred over the other colors
B) That there is no preference between the colors
C) That any one color is preferred over the other colors
D) Impossible to determine
A) That willow green is preferred over the other colors
B) That there is no preference between the colors
C) That any one color is preferred over the other colors
D) Impossible to determine
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75
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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76
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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77
When determining how well an observed set of frequencies fit an expected set of frequencies, what is the test statistic?
A) F-statistic
B) t-statistic.
C)
D) z-statistic
A) F-statistic
B) t-statistic.
C)
D) z-statistic
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78
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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79
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 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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80
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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