Deck 15: Chi-Square Tests

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Question
In a chi-square test of a 5 × 5 contingency table at α = .05, the critical value is 37.65.
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Question
In a chi-square test for independence, observed frequencies must be at least 5 in every cell.
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The chi-square test is based on the analysis of frequencies.
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In a hypothesis test using chi-square, if the null hypothesis is true, the sample value of the sample chi-square test statistic will be exactly zero.
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The chi-square test can only be used to assess independence between two variables.
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In samples drawn from a population in which the row and column categories are independent, the value of the chi-square test statistic will be zero.
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In a chi-square test for independence, expected frequencies must be integers (or rounded to the nearest integer).
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A chi-square distribution is always skewed right.
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A large negative chi-square test statistic would indicate that the null hypothesis should be rejected.
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The chi-square test is unreliable when there are any cells with small observed frequency counts.
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The chi-square test for independence is a nonparametric test (no parameters are estimated).
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If two variables are independent, we would anticipate a chi-square test statistic close to zero.
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In a chi-square test for independence, observed and expected frequencies must sum across to the same row totals and down to the same column totals.
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Observed frequencies in a chi-square goodness-of-fit test for normality may be less than 5, or even 0, in some cells, as long as the expected frequencies are large enough.
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A chi-square test for independence is called a distribution-free test since the test is based on categorical data rather than on populations that follow any particular distribution.
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The shape of the chi-square distribution depends only on its degrees of freedom.
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Cochran's Rule requires observed frequencies of 5 or more in each cell of a contingency table.
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The degrees of freedom in a 3 × 4 chi-square contingency table would equal 11.
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The null hypothesis for a chi-square contingency test of independence for two variables always assumes that the variables are independent.
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The null hypothesis for a chi-square test on a contingency table is that the variables are dependent.
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For a chi-square goodness-of-fit test for a uniform distribution with 7 categories, we would use the critical value for 6 degrees of freedom.
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An attraction of the Kolmogorov-Smirnov test is that it is fairly easy to do without a computer.
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In a goodness-of-fit test, a linear probability plot suggests that the null hypothesis should be rejected.
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In an ECDF test for goodness-of-fit, the n observations are grouped into categories rather than being treated individually.
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The Anderson-Darling test is used to test the assumption of normality.
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In a chi-square goodness-of-fit test, a small p-value would indicate a good fit to the hypothesized distribution.
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ECDF tests have an advantage over the chi-square goodness-of-fit test on frequencies because an ECDF test treats observations individually.
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Probability plots are used to test the assumption of normality.
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For a chi-square goodness-of-fit test for a normal distribution using 7 categories with estimated mean and standard deviation, we would use the critical value for 4 degrees of freedom.
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A probability plot usually allows outliers to be detected.
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For a chi-square goodness-of-fit test for a normal distribution using 8 categories with estimated mean and standard deviation, we would use the critical value for 7 degrees of freedom.
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The Poisson goodness-of-fit test is inappropriate for continuous data.
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In a chi-square goodness-of-fit test, we lose one degree of freedom for each parameter estimated.
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The Kolmogorov-Smirnov and Anderson-Darling tests are based on the ECDF (Empirical Cumulative Distribution Function).
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When raw data are available, ECDF tests usually surpass the chi-square test in their ability to detect departures from the distribution specified in the null hypothesis.
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In a chi-square goodness-of-fit test, a sample of n observations has n - 1 degrees of freedom.
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An attraction of the Anderson-Darling test is that it is fairly easy to do without a computer.
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Goodness-of-fit tests using the ECDF (Empirical Cumulative Distribution Function) compare the actual cumulative frequencies with expected cumulative frequencies for each observation under the assumption that the data came from the hypothesized distribution.
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For a chi-square goodness-of-fit test for a uniform distribution with 5 categories, we would use the critical value for 4 degrees of freedom.
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In a chi-square goodness-of-fit test, we gain one degree of freedom if n increases by 1.
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If samples are drawn from a population that is normal, a goodness-of-fit test for normality could yield:

A)Type I error but not Type II error.
B)Type II error but not Type I error.
C)Either Type I error or Type II error.
D)Both Type I and Type II errors.
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A chi-square test of independence is a one-tailed test. The reason is that:

A)we are testing whether the frequencies exceed their expected values.
B)we square the deviations, so the test statistic lies at or above zero.
C)hypothesis tests are one-tailed tests when dealing with sample data.
D)the chi-square distribution is positively skewed.
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In a chi-square test of independence, the number of degrees of freedom equals the:

A)number of observations minus one.
B)number of categories minus one.
C)number of rows minus one times the number of columns minus one.
D)number of sample observations minus the missing observations.
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To determine how well an observed set of frequencies fits an expected set of frequencies from a Poisson distribution, we must estimate:

A)no parameters.
B)one parameter (λ).
C)two parameters (μ, σ).
D)three parameters (μ, σ, n).
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Which of these statements concerning a chi-square goodness-of-fit test is correct?

A)Data could be ratio or interval measurements.
B)Population must be normally distributed.
C)All the expected frequencies must be equal.
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We sometimes combine two row or column categories in a chi-square test when:

A)observed frequencies are more than 5.
B)observed frequencies are less than 5.
C)expected frequencies are more than 5.
D)expected frequencies are less than 5.
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In a test for a uniform distribution with k categories, the expected frequency is n/k in each cell.
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The number of cars waiting at a certain residential neighborhood stoplight is observed at 6:00 a.m. on 160 different days. The observed sample frequencies are shown here:  Number of cars waiting 01236060503020\begin{array} { | l | r | r | r | r | } \hline \text { Number of cars waiting } & 0 & 1 & 2 & 3 \\\hline 60 & 60 & 50 & 30 & 20 \\\hline\end{array} Under the null hypothesis of a uniform distribution, the expected number of days we would see 0 cars is:

A)10.
B)20.
C)30.
D)40.
Question
Kortholts that fail to meet certain precise specifications must be reworked on the next day until they are within the desired specifications. A sample of one day's output of kortholts from the Melodic Kortholt Company showed the following frequencies:  Plant A  Plant B  Row Total Specification Met18585270 Specification Not Met 151530 Col Total 200100300\begin{array}{|c|c|c|c|}\hline&\text { Plant A }&\text { Plant B }&\text { Row Total }\\\hline \text {Specification Met}&185 & 85&270 \\\hline \text { Specification Not Met }&15 & 15& 30\\\hline \text { Col Total }&200 & 100 &300\\\hline\end{array} Find the chi-square test statistic for a hypothesis of independence.

A)7.22
B)4.17
C)5.13
D)6.08
Question
Which of the following is not a potential solution to the problem that arises when not all expected frequencies are 5 or more in a chi-square test for independence?

A)Combine some of the columns
B)Combine some of the rows
C)Increase the sample size
D)Add more rows or columns
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The critical value in a chi-square test for independence depends on:

A)the normality of the data.
B)the variance of the data.
C)the number of categories.
D)the expected frequencies.
Question
Kortholts that fail to meet certain precise specifications must be reworked on the next day until they are within the desired specifications. A sample of one day's output of kortholts from the Melodic Kortholt Company showed the following frequencies:  Plant A  Plant B  Row Total Specification Met18585270 Specification Not Met 151530 Col Total 200100300\begin{array}{|c|c|c|c|}\hline&\text { Plant A }&\text { Plant B }&\text { Row Total }\\\hline \text {Specification Met}&185 & 85&270 \\\hline \text { Specification Not Met }&15 & 15& 30\\\hline \text { Col Total }&200 & 100 &300\\\hline\end{array} Find the p-value for the chi-square test statistic for a hypothesis of independence.

A)Less than .01
B)Between .01 and .025
C)Between .025 and .05
D)Greater than .05
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A chi-square goodness of fit test for a normal distribution used 40 observations, and the mean and standard deviation were estimated from the sample. The test used six categories. We would use how many degrees of freedom in looking up the critical value for the test?

A)39
B)37
C)5
D)3
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A chi-square goodness of fit test for a normal distribution used 60 observations, and the mean and standard deviation were estimated from the sample. The test used seven categories. We would use how many degrees of freedom in looking up the critical value for the test?

A)6
B)4
C)59
D)57
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A proofreader checked 160 ads for grammatical errors. The sample frequency distribution is shown:  Number of Errors 0123 Observed Frequency 10657114\begin{array} { | l | c | c | c | c | } \hline \text { Number of Errors } & 0 & 1 & 2 & 3 \\\hline \text { Observed Frequency } & 10 & 65 & 71 & 14 \\\hline\end{array} Using a goodness-of-fit test to determine whether this distribution is uniform would result in a chi-square test statistic of approximately:

A)55.
B)79.
C)85.
D) 161.
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A proofreader checked 160 ads for grammatical errors. The distribution obtained is shown:  Number of Errors 0123 Observed Frequency 10657114\begin{array} { | l | c | c | c | c | } \hline \text { Number of Errors } & 0 & 1 & 2 & 3 \\\hline \text { Observed Frequency } & 10 & 65 & 71 & 14 \\\hline\end{array} At ? = .01, what decision would we reach in a goodness-of-fit test to see whether this sample came from a uniform distribution?

A)Reject the null and conclude the distribution is not uniform.
B)Conclude that there is insufficient evidence to reject the null.
C)No conclusion can be made due to small expected frequencies.
D)No conclusion can be made due to inadequate sample size.
Question
In order to apply the chi-square test of independence, we prefer to have:

A)at least 5 observed frequencies in each cell.
B)at least 5 expected observations in each cell.
C)at least 5 percent of the observations in each cell.
D)not more than 5 observations in each cell.
Question
Which of these statements concerning a chi-square goodness-of-fit test is correct?

A)It is inapplicable to test for a normal distribution with open-ended top and bottom classes.
B)It is generally a better test than the chi-square test of independence.
C)There is no way to get the degrees of freedom since the right tail goes to infinity.
D)It can be used to test whether a sample follows a specified distribution.
Question
An operations analyst counted the number of arrivals per minute at a bank ATM in each of 30 randomly chosen minutes. The results were: 0, 3, 3, 2, 1, 0, 1, 0, 0, 1, 1, 1, 2, 1, 0, 1, 0, 1, 2, 1, 1, 2, 1, 0, 1, 2, 0, 1, 0, 1. Which goodness-of-fit test would you recommend?

A)Uniform
B)Poisson
C)Normal
D)Binomial
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A proofreader checked 160 ads for grammatical errors. The sample frequency distribution is shown:  Number of Errors 0123 Observed Frequency 10657114\begin{array} { | l | c | c | c | c | } \hline \text { Number of Errors } & 0 & 1 & 2 & 3 \\\hline \text { Observed Frequency } & 10 & 65 & 71 & 14 \\\hline\end{array} Under the null hypothesis of a uniform distribution, the expected number of times we would get 0 errors is:

A)10.
B)20.
C)30.
D)40.
Question
To carry out a chi-square goodness-of-fit test for normality, you need at least:

A)5 categories altogether.
B)5 observations in each category.
C)5 expected observations in each category.
D)50 samples or more.
Question
As an independent project, a team of statistics students tabulated the types of vehicles that were parked in four different suburban shopping malls.                           MrN Location \text { MrN Location }
 Vehicle Type  Somerset  Oakland  Great Lakes  Jamestown  Row Tolal  Car 44493664193Minivan 2115181367 Full-size Van 233210 SUN 1927261284Truck 13617946 Col Total100100100100400\begin{array}{|c|c|c|c|c|c|}\hline\text { Vehicle Type }& \text { Somerset } & \text { Oakland } & \text { Great Lakes } & \text { Jamestown }&\text { Row Tolal }\\\hline \text { Car }&44 & 49 & 36 & 64&193 \\\hline \text {Minivan }&21 & 15 & 18 & 13 &67\\\hline \text { Full-size Van }&2 & 3 & 3 & 2 &10\\\hline \text { SUN }&19 & 27 & 26 & 12&84 \\\hline \text {Truck }&13 & 6 & 17 & 9 &46\\\hline \text { Col Total}&100 & 100 & 100 & 100 &400\\\hline\end{array} For a chi-square test of independence, degrees of freedom would be:

A)20.
B)12.
C)399.
D)6.
Question
As an independent project, a team of statistics students tabulated the types of vehicles that were parked in four different suburban shopping malls.                           MrN Location \text { MrN Location }
 Vehicle Type  Somerset  Oakland  Great Lakes  Jamestown  Row Tolal  Car 44493664193Minivan 2115181367 Full-size Van 233210 SUN 1927261284Truck 13617946 Col Total100100100100400\begin{array}{|c|c|c|c|c|c|}\hline\text { Vehicle Type }& \text { Somerset } & \text { Oakland } & \text { Great Lakes } & \text { Jamestown }&\text { Row Tolal }\\\hline \text { Car }&44 & 49 & 36 & 64&193 \\\hline \text {Minivan }&21 & 15 & 18 & 13 &67\\\hline \text { Full-size Van }&2 & 3 & 3 & 2 &10\\\hline \text { SUN }&19 & 27 & 26 & 12&84 \\\hline \text {Truck }&13 & 6 & 17 & 9 &46\\\hline \text { Col Total}&100 & 100 & 100 & 100 &400\\\hline\end{array} For a chi-square test of independence, the critical value for ? = .10 is:

A)10.64.
B)14.68.
C)28.41.
D)18.55.
Question
The table below is a tabulation of opinions of employees of Axolotl Corporation, who were sampled at random from pay records and asked to complete an anonymous job satisfaction survey.                      Degree of Satisfaction  \text { Degree of Satisfaction }
 Pay Type  Satisfied  Neutral  Dissatisfied  Row Total  Salaried 40101060 Hourly 805050180 Col Total 1206060240\begin{array} { |c | c | c | c | c | } \hline{ \text { Pay Type } } & \text { Satisfied } & { \text { Neutral } } & \text { Dissatisfied } & \text { Row Total } \\\hline \text { Salaried } & 40 & 10 & 10 & 60 \\\hline\text { Hourly } & 80 & 50 & 50 & 180 \\\hline \text { Col Total } & 120 & 60 & 60 & 240\\\hline\end{array} For a chi-square test of independence, degrees of freedom would be:

A)2
B)3
C)4
D)6
Question
As an independent project, a team of statistics students tabulated the types of vehicles that were parked in four different suburban shopping malls.                           MrN Location \text { MrN Location }
 Vehicle Type  Somerset  Oakland  Great Lakes  Jamestown  Row Tolal  Car 44493664193Minivan 2115181367 Full-size Van 233210 SUN 1927261284Truck 13617946 Col Total100100100100400\begin{array}{|c|c|c|c|c|c|}\hline\text { Vehicle Type }& \text { Somerset } & \text { Oakland } & \text { Great Lakes } & \text { Jamestown }&\text { Row Tolal }\\\hline \text { Car }&44 & 49 & 36 & 64&193 \\\hline \text {Minivan }&21 & 15 & 18 & 13 &67\\\hline \text { Full-size Van }&2 & 3 & 3 & 2 &10\\\hline \text { SUN }&19 & 27 & 26 & 12&84 \\\hline \text {Truck }&13 & 6 & 17 & 9 &46\\\hline \text { Col Total}&100 & 100 & 100 & 100 &400\\\hline\end{array} Assuming independence, the expected frequency of SUVs in Jamestown is:

A)12.
B)21.
C)75.
D)60.
Question
Employees of OxCo Mfg. were surveyed to evaluate the company's pension plan. The table below displays some of the results of the survey. <strong>Employees of OxCo Mfg. were surveyed to evaluate the company's pension plan. The table below displays some of the results of the survey.   The expected frequency for the shaded cell in the table would be:</strong> A)163. B)158. C)165. D) 160. <div style=padding-top: 35px> The expected frequency for the shaded cell in the table would be:

A)163.
B)158.
C)165.
D) 160.
Question
The table below is a tabulation of opinions of employees of Axolotl Corporation, who were sampled at random from pay records and asked to complete an anonymous job satisfaction survey.                      Degree of Satisfaction  \text { Degree of Satisfaction }
 Pay Type  Satisfied  Neutral  Dissatisfied  Row Total  Salaried 40101060 Hourly 805050180 Col Total 1206060240\begin{array} { |c | c | c | c | c | } \hline{ \text { Pay Type } } & \text { Satisfied } & { \text { Neutral } } & \text { Dissatisfied } & \text { Row Total } \\\hline \text { Salaried } & 40 & 10 & 10 & 60 \\\hline\text { Hourly } & 80 & 50 & 50 & 180 \\\hline \text { Col Total } & 120 & 60 & 60 & 240\\\hline\end{array} Assuming independence, the expected frequency of satisfied hourly employees is:

A)80.
B)90.
C)75.
D)60.
Question
An operations analyst counted the number of arrivals per minute at an ATM in each of 30 randomly chosen minutes. The results were: 0, 3, 3, 2, 1, 0, 1, 0, 0, 1, 1, 1, 2, 1, 0, 1, 0, 1, 2, 1, 1, 2, 1, 0, 1, 2, 0, 1, 0, 1. For the Poisson goodness-of-fit test, what is the expected frequency of the data value X = 1?

A)Impossible to determine.
B)11.04
C)1.00
D)2.47
Question
The table below is a tabulation of opinions of employees of Axolotl Corporation, who were sampled at random from pay records and asked to complete an anonymous job satisfaction survey.                      Degree of Satisfaction  \text { Degree of Satisfaction }
 Pay Type  Satisfied  Neutral  Dissatisfied  Row Total  Salaried 40101060 Hourly 805050180 Col Total 1206060240\begin{array} { |c | c | c | c | c | } \hline{ \text { Pay Type } } & \text { Satisfied } & { \text { Neutral } } & \text { Dissatisfied } & \text { Row Total } \\\hline \text { Salaried } & 40 & 10 & 10 & 60 \\\hline\text { Hourly } & 80 & 50 & 50 & 180 \\\hline \text { Col Total } & 120 & 60 & 60 & 240\\\hline\end{array} For a chi-square test of independence, the critical value for ? = .01 is:

A)9.210.
B)4.605.
C)11.34.
D)16.81.
Question
You test a hypothesis of independence of two variables. The number of observations is 500 and you have classified the data into a 4 by 4 contingency table. The test statistic has __________ degrees of freedom.

A)16
B)9
C)499
D)498
Question
Students in an introductory college economics class were asked how many credits they had earned in college, and how certain they were about their choice of major. Their replies are summarized below.                           Degree of Certainty  \text { Degree of Certainty }
 Credits Earned  Very Uncertain  Somewhat Certain  Very Certain Row Tot Under 10 1416646 10 through 591682044 60 or more 2142238 Col Total 423848128\begin{array} { |c | c | c | c | c | } \hline \text { Credits Earned }&{ \text { Very Uncertain } } & \text { Somewhat Certain } & { \text { Very Certain} } & \text { Row Tot } \\\hline \text {Under 10 } & 14 & 16 & 6 &46 \\\hline\text { 10 through 59} & 16& 8 & 20 & 44 \\\hline \text { 60 or more } & 2 & 14 & 22 & 38\\\hline \text { Col Total } & 42 & 38 & 48 & 128\\\hline\end{array} Assuming independence, the expected frequency of very uncertain students with 60 credits or more is:

A)12.47.
B)2.00.
C)14.56.
D)11.09.
Question
You want to test the hypothesis that the prime rate and inflation are independent. The following table is prepared for the test on the basis of the results of a random sample, collected in various countries and various time periods: <strong>You want to test the hypothesis that the prime rate and inflation are independent. The following table is prepared for the test on the basis of the results of a random sample, collected in various countries and various time periods:   Using α = .05, what is the critical value of the test statistic that you would use?</strong> A)3.841 B)12.59 C)5.991 D)7.815 <div style=padding-top: 35px> Using α = .05, what is the critical value of the test statistic that you would use?

A)3.841
B)12.59
C)5.991
D)7.815
Question
Students in an introductory college economics class were asked how many credits they had earned in college, and how certain they were about their choice of major. Their replies are summarized below.                           Degree of Certainty  \text { Degree of Certainty }
 Credits Earned  Very Uncertain  Somewhat Certain  Very Certain Row Tot Under 10 1416646 10 through 591682044 60 or more 2142238 Col Total 423848128\begin{array} { |c | c | c | c | c | } \hline \text { Credits Earned }&{ \text { Very Uncertain } } & \text { Somewhat Certain } & { \text { Very Certain} } & \text { Row Tot } \\\hline \text {Under 10 } & 14 & 16 & 6 &46 \\\hline\text { 10 through 59} & 16& 8 & 20 & 44 \\\hline \text { 60 or more } & 2 & 14 & 22 & 38\\\hline \text { Col Total } & 42 & 38 & 48 & 128\\\hline\end{array} For a chi-square test of independence, degrees of freedom would be:

A)2.
B)9.
C)4.
D)127.
Question
Students in an introductory college economics class were asked how many credits they had earned in college, and how certain they were about their choice of major. Their replies are summarized below.                           Degree of Certainty  \text { Degree of Certainty }
 Credits Earned  Very Uncertain  Somewhat Certain  Very Certain Row Tot Under 10 1416646 10 through 591682044 60 or more 2142238 Col Total 423848128\begin{array} { |c | c | c | c | c | } \hline \text { Credits Earned }&{ \text { Very Uncertain } } & \text { Somewhat Certain } & { \text { Very Certain} } & \text { Row Tot } \\\hline \text {Under 10 } & 14 & 16 & 6 &46 \\\hline\text { 10 through 59} & 16& 8 & 20 & 44 \\\hline \text { 60 or more } & 2 & 14 & 22 & 38\\\hline \text { Col Total } & 42 & 38 & 48 & 128\\\hline\end{array} Under the assumption of independence, the expected frequency in the upper left cell is:

A)15.09.
B)24.00.
C)19.72.
D)20.22.
Question
Employees of OxCo Mfg. were surveyed to evaluate the company's pension plan. The table below displays some of the results of the survey. <strong>Employees of OxCo Mfg. were surveyed to evaluate the company's pension plan. The table below displays some of the results of the survey.   The appropriate conclusion would be:</strong> A)do not reject H<sub>0</sub>. B)reject H<sub>0</sub> at α = .10. C)reject H<sub>0</sub> at α = .05. D)reject H<sub>0</sub> at α = .01. <div style=padding-top: 35px> The appropriate conclusion would be:

A)do not reject H0.
B)reject H0 at α = .10.
C)reject H0 at α = .05.
D)reject H0 at α = .01.
Question
Employees of OxCo Mfg. were surveyed to evaluate the company's pension plan. The table below displays some of the results of the survey. <strong>Employees of OxCo Mfg. were surveyed to evaluate the company's pension plan. The table below displays some of the results of the survey.   Degrees of freedom for this test (shaded cell below the table) would be:</strong> A)6. B)7. C)799. D)12. <div style=padding-top: 35px> Degrees of freedom for this test (shaded cell below the table) would be:

A)6.
B)7.
C)799.
D)12.
Question
You want to test the hypothesis that the prime rate and inflation are independent. The following table is prepared for the test on the basis of the results of a random sample, collected in various countries and various time periods: <strong>You want to test the hypothesis that the prime rate and inflation are independent. The following table is prepared for the test on the basis of the results of a random sample, collected in various countries and various time periods:   The expected frequency for the shaded cell is:</strong> A)22.5. B)30. C)40. D)40.5. <div style=padding-top: 35px> The expected frequency for the shaded cell is:

A)22.5.
B)30.
C)40.
D)40.5.
Question
You want to test the hypothesis that the prime rate and inflation are independent. The following table is prepared for the test on the basis of the results of a random sample, collected in various countries and various time periods: <strong>You want to test the hypothesis that the prime rate and inflation are independent. The following table is prepared for the test on the basis of the results of a random sample, collected in various countries and various time periods:   What is the value of the test statistic?</strong> A)306.25 B)0.00 C)54.44 D)13.61 <div style=padding-top: 35px> What is the value of the test statistic?

A)306.25
B)0.00
C)54.44
D)13.61
Question
Students in an introductory college economics class were asked how many credits they had earned in college, and how certain they were about their choice of major. Their replies are summarized below.                           Degree of Certainty  \text { Degree of Certainty }
 Credits Earned  Very Uncertain  Somewhat Certain  Very Certain Row Tot Under 10 1416646 10 through 591682044 60 or more 2142238 Col Total 423848128\begin{array} { |c | c | c | c | c | } \hline \text { Credits Earned }&{ \text { Very Uncertain } } & \text { Somewhat Certain } & { \text { Very Certain} } & \text { Row Tot } \\\hline \text {Under 10 } & 14 & 16 & 6 &46 \\\hline\text { 10 through 59} & 16& 8 & 20 & 44 \\\hline \text { 60 or more } & 2 & 14 & 22 & 38\\\hline \text { Col Total } & 42 & 38 & 48 & 128\\\hline\end{array} For a chi-square test of independence, the critical value for ? = .05 is:

A)5.991.
B)7.815.
C)9.488.
D)16.92.
Question
Students in an introductory college economics class were asked how many credits they had earned in college, and how certain they were about their choice of major. Their replies are summarized below.                           Degree of Certainty  \text { Degree of Certainty }
 Credits Earned  Very Uncertain  Somewhat Certain  Very Certain Row Tot Under 10 1416646 10 through 591682044 60 or more 2142238 Col Total 423848128\begin{array} { |c | c | c | c | c | } \hline \text { Credits Earned }&{ \text { Very Uncertain } } & \text { Somewhat Certain } & { \text { Very Certain} } & \text { Row Tot } \\\hline \text {Under 10 } & 14 & 16 & 6 &46 \\\hline\text { 10 through 59} & 16& 8 & 20 & 44 \\\hline \text { 60 or more } & 2 & 14 & 22 & 38\\\hline \text { Col Total } & 42 & 38 & 48 & 128\\\hline\end{array} Which statement is most nearly correct?

A)The contingency table violates Cochran's Rule.
B)Visual inspection of column frequencies suggests independence.
C)At ? = .05 we would easily reject the null hypothesis of independence.
D)At ? = .05 we cannot reject he null hypothesis of independence.
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Deck 15: Chi-Square Tests
1
In a chi-square test of a 5 × 5 contingency table at α = .05, the critical value is 37.65.
False
2
In a chi-square test for independence, observed frequencies must be at least 5 in every cell.
False
3
The chi-square test is based on the analysis of frequencies.
True
4
In a hypothesis test using chi-square, if the null hypothesis is true, the sample value of the sample chi-square test statistic will be exactly zero.
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5
The chi-square test can only be used to assess independence between two variables.
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6
In samples drawn from a population in which the row and column categories are independent, the value of the chi-square test statistic will be zero.
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7
In a chi-square test for independence, expected frequencies must be integers (or rounded to the nearest integer).
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8
A chi-square distribution is always skewed right.
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9
A large negative chi-square test statistic would indicate that the null hypothesis should be rejected.
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10
The chi-square test is unreliable when there are any cells with small observed frequency counts.
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11
The chi-square test for independence is a nonparametric test (no parameters are estimated).
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12
If two variables are independent, we would anticipate a chi-square test statistic close to zero.
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13
In a chi-square test for independence, observed and expected frequencies must sum across to the same row totals and down to the same column totals.
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14
Observed frequencies in a chi-square goodness-of-fit test for normality may be less than 5, or even 0, in some cells, as long as the expected frequencies are large enough.
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15
A chi-square test for independence is called a distribution-free test since the test is based on categorical data rather than on populations that follow any particular distribution.
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16
The shape of the chi-square distribution depends only on its degrees of freedom.
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17
Cochran's Rule requires observed frequencies of 5 or more in each cell of a contingency table.
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18
The degrees of freedom in a 3 × 4 chi-square contingency table would equal 11.
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19
The null hypothesis for a chi-square contingency test of independence for two variables always assumes that the variables are independent.
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20
The null hypothesis for a chi-square test on a contingency table is that the variables are dependent.
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21
For a chi-square goodness-of-fit test for a uniform distribution with 7 categories, we would use the critical value for 6 degrees of freedom.
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22
An attraction of the Kolmogorov-Smirnov test is that it is fairly easy to do without a computer.
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23
In a goodness-of-fit test, a linear probability plot suggests that the null hypothesis should be rejected.
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24
In an ECDF test for goodness-of-fit, the n observations are grouped into categories rather than being treated individually.
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25
The Anderson-Darling test is used to test the assumption of normality.
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26
In a chi-square goodness-of-fit test, a small p-value would indicate a good fit to the hypothesized distribution.
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27
ECDF tests have an advantage over the chi-square goodness-of-fit test on frequencies because an ECDF test treats observations individually.
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28
Probability plots are used to test the assumption of normality.
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29
For a chi-square goodness-of-fit test for a normal distribution using 7 categories with estimated mean and standard deviation, we would use the critical value for 4 degrees of freedom.
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30
A probability plot usually allows outliers to be detected.
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31
For a chi-square goodness-of-fit test for a normal distribution using 8 categories with estimated mean and standard deviation, we would use the critical value for 7 degrees of freedom.
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32
The Poisson goodness-of-fit test is inappropriate for continuous data.
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33
In a chi-square goodness-of-fit test, we lose one degree of freedom for each parameter estimated.
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34
The Kolmogorov-Smirnov and Anderson-Darling tests are based on the ECDF (Empirical Cumulative Distribution Function).
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35
When raw data are available, ECDF tests usually surpass the chi-square test in their ability to detect departures from the distribution specified in the null hypothesis.
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36
In a chi-square goodness-of-fit test, a sample of n observations has n - 1 degrees of freedom.
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37
An attraction of the Anderson-Darling test is that it is fairly easy to do without a computer.
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38
Goodness-of-fit tests using the ECDF (Empirical Cumulative Distribution Function) compare the actual cumulative frequencies with expected cumulative frequencies for each observation under the assumption that the data came from the hypothesized distribution.
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39
For a chi-square goodness-of-fit test for a uniform distribution with 5 categories, we would use the critical value for 4 degrees of freedom.
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40
In a chi-square goodness-of-fit test, we gain one degree of freedom if n increases by 1.
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41
If samples are drawn from a population that is normal, a goodness-of-fit test for normality could yield:

A)Type I error but not Type II error.
B)Type II error but not Type I error.
C)Either Type I error or Type II error.
D)Both Type I and Type II errors.
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42
A chi-square test of independence is a one-tailed test. The reason is that:

A)we are testing whether the frequencies exceed their expected values.
B)we square the deviations, so the test statistic lies at or above zero.
C)hypothesis tests are one-tailed tests when dealing with sample data.
D)the chi-square distribution is positively skewed.
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43
In a chi-square test of independence, the number of degrees of freedom equals the:

A)number of observations minus one.
B)number of categories minus one.
C)number of rows minus one times the number of columns minus one.
D)number of sample observations minus the missing observations.
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44
To determine how well an observed set of frequencies fits an expected set of frequencies from a Poisson distribution, we must estimate:

A)no parameters.
B)one parameter (λ).
C)two parameters (μ, σ).
D)three parameters (μ, σ, n).
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45
Which of these statements concerning a chi-square goodness-of-fit test is correct?

A)Data could be ratio or interval measurements.
B)Population must be normally distributed.
C)All the expected frequencies must be equal.
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46
We sometimes combine two row or column categories in a chi-square test when:

A)observed frequencies are more than 5.
B)observed frequencies are less than 5.
C)expected frequencies are more than 5.
D)expected frequencies are less than 5.
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47
In a test for a uniform distribution with k categories, the expected frequency is n/k in each cell.
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48
The number of cars waiting at a certain residential neighborhood stoplight is observed at 6:00 a.m. on 160 different days. The observed sample frequencies are shown here:  Number of cars waiting 01236060503020\begin{array} { | l | r | r | r | r | } \hline \text { Number of cars waiting } & 0 & 1 & 2 & 3 \\\hline 60 & 60 & 50 & 30 & 20 \\\hline\end{array} Under the null hypothesis of a uniform distribution, the expected number of days we would see 0 cars is:

A)10.
B)20.
C)30.
D)40.
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49
Kortholts that fail to meet certain precise specifications must be reworked on the next day until they are within the desired specifications. A sample of one day's output of kortholts from the Melodic Kortholt Company showed the following frequencies:  Plant A  Plant B  Row Total Specification Met18585270 Specification Not Met 151530 Col Total 200100300\begin{array}{|c|c|c|c|}\hline&\text { Plant A }&\text { Plant B }&\text { Row Total }\\\hline \text {Specification Met}&185 & 85&270 \\\hline \text { Specification Not Met }&15 & 15& 30\\\hline \text { Col Total }&200 & 100 &300\\\hline\end{array} Find the chi-square test statistic for a hypothesis of independence.

A)7.22
B)4.17
C)5.13
D)6.08
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50
Which of the following is not a potential solution to the problem that arises when not all expected frequencies are 5 or more in a chi-square test for independence?

A)Combine some of the columns
B)Combine some of the rows
C)Increase the sample size
D)Add more rows or columns
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51
The critical value in a chi-square test for independence depends on:

A)the normality of the data.
B)the variance of the data.
C)the number of categories.
D)the expected frequencies.
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52
Kortholts that fail to meet certain precise specifications must be reworked on the next day until they are within the desired specifications. A sample of one day's output of kortholts from the Melodic Kortholt Company showed the following frequencies:  Plant A  Plant B  Row Total Specification Met18585270 Specification Not Met 151530 Col Total 200100300\begin{array}{|c|c|c|c|}\hline&\text { Plant A }&\text { Plant B }&\text { Row Total }\\\hline \text {Specification Met}&185 & 85&270 \\\hline \text { Specification Not Met }&15 & 15& 30\\\hline \text { Col Total }&200 & 100 &300\\\hline\end{array} Find the p-value for the chi-square test statistic for a hypothesis of independence.

A)Less than .01
B)Between .01 and .025
C)Between .025 and .05
D)Greater than .05
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53
A chi-square goodness of fit test for a normal distribution used 40 observations, and the mean and standard deviation were estimated from the sample. The test used six categories. We would use how many degrees of freedom in looking up the critical value for the test?

A)39
B)37
C)5
D)3
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54
A chi-square goodness of fit test for a normal distribution used 60 observations, and the mean and standard deviation were estimated from the sample. The test used seven categories. We would use how many degrees of freedom in looking up the critical value for the test?

A)6
B)4
C)59
D)57
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55
A proofreader checked 160 ads for grammatical errors. The sample frequency distribution is shown:  Number of Errors 0123 Observed Frequency 10657114\begin{array} { | l | c | c | c | c | } \hline \text { Number of Errors } & 0 & 1 & 2 & 3 \\\hline \text { Observed Frequency } & 10 & 65 & 71 & 14 \\\hline\end{array} Using a goodness-of-fit test to determine whether this distribution is uniform would result in a chi-square test statistic of approximately:

A)55.
B)79.
C)85.
D) 161.
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56
A proofreader checked 160 ads for grammatical errors. The distribution obtained is shown:  Number of Errors 0123 Observed Frequency 10657114\begin{array} { | l | c | c | c | c | } \hline \text { Number of Errors } & 0 & 1 & 2 & 3 \\\hline \text { Observed Frequency } & 10 & 65 & 71 & 14 \\\hline\end{array} At ? = .01, what decision would we reach in a goodness-of-fit test to see whether this sample came from a uniform distribution?

A)Reject the null and conclude the distribution is not uniform.
B)Conclude that there is insufficient evidence to reject the null.
C)No conclusion can be made due to small expected frequencies.
D)No conclusion can be made due to inadequate sample size.
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57
In order to apply the chi-square test of independence, we prefer to have:

A)at least 5 observed frequencies in each cell.
B)at least 5 expected observations in each cell.
C)at least 5 percent of the observations in each cell.
D)not more than 5 observations in each cell.
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58
Which of these statements concerning a chi-square goodness-of-fit test is correct?

A)It is inapplicable to test for a normal distribution with open-ended top and bottom classes.
B)It is generally a better test than the chi-square test of independence.
C)There is no way to get the degrees of freedom since the right tail goes to infinity.
D)It can be used to test whether a sample follows a specified distribution.
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59
An operations analyst counted the number of arrivals per minute at a bank ATM in each of 30 randomly chosen minutes. The results were: 0, 3, 3, 2, 1, 0, 1, 0, 0, 1, 1, 1, 2, 1, 0, 1, 0, 1, 2, 1, 1, 2, 1, 0, 1, 2, 0, 1, 0, 1. Which goodness-of-fit test would you recommend?

A)Uniform
B)Poisson
C)Normal
D)Binomial
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60
A proofreader checked 160 ads for grammatical errors. The sample frequency distribution is shown:  Number of Errors 0123 Observed Frequency 10657114\begin{array} { | l | c | c | c | c | } \hline \text { Number of Errors } & 0 & 1 & 2 & 3 \\\hline \text { Observed Frequency } & 10 & 65 & 71 & 14 \\\hline\end{array} Under the null hypothesis of a uniform distribution, the expected number of times we would get 0 errors is:

A)10.
B)20.
C)30.
D)40.
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61
To carry out a chi-square goodness-of-fit test for normality, you need at least:

A)5 categories altogether.
B)5 observations in each category.
C)5 expected observations in each category.
D)50 samples or more.
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62
As an independent project, a team of statistics students tabulated the types of vehicles that were parked in four different suburban shopping malls.                           MrN Location \text { MrN Location }
 Vehicle Type  Somerset  Oakland  Great Lakes  Jamestown  Row Tolal  Car 44493664193Minivan 2115181367 Full-size Van 233210 SUN 1927261284Truck 13617946 Col Total100100100100400\begin{array}{|c|c|c|c|c|c|}\hline\text { Vehicle Type }& \text { Somerset } & \text { Oakland } & \text { Great Lakes } & \text { Jamestown }&\text { Row Tolal }\\\hline \text { Car }&44 & 49 & 36 & 64&193 \\\hline \text {Minivan }&21 & 15 & 18 & 13 &67\\\hline \text { Full-size Van }&2 & 3 & 3 & 2 &10\\\hline \text { SUN }&19 & 27 & 26 & 12&84 \\\hline \text {Truck }&13 & 6 & 17 & 9 &46\\\hline \text { Col Total}&100 & 100 & 100 & 100 &400\\\hline\end{array} For a chi-square test of independence, degrees of freedom would be:

A)20.
B)12.
C)399.
D)6.
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63
As an independent project, a team of statistics students tabulated the types of vehicles that were parked in four different suburban shopping malls.                           MrN Location \text { MrN Location }
 Vehicle Type  Somerset  Oakland  Great Lakes  Jamestown  Row Tolal  Car 44493664193Minivan 2115181367 Full-size Van 233210 SUN 1927261284Truck 13617946 Col Total100100100100400\begin{array}{|c|c|c|c|c|c|}\hline\text { Vehicle Type }& \text { Somerset } & \text { Oakland } & \text { Great Lakes } & \text { Jamestown }&\text { Row Tolal }\\\hline \text { Car }&44 & 49 & 36 & 64&193 \\\hline \text {Minivan }&21 & 15 & 18 & 13 &67\\\hline \text { Full-size Van }&2 & 3 & 3 & 2 &10\\\hline \text { SUN }&19 & 27 & 26 & 12&84 \\\hline \text {Truck }&13 & 6 & 17 & 9 &46\\\hline \text { Col Total}&100 & 100 & 100 & 100 &400\\\hline\end{array} For a chi-square test of independence, the critical value for ? = .10 is:

A)10.64.
B)14.68.
C)28.41.
D)18.55.
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64
The table below is a tabulation of opinions of employees of Axolotl Corporation, who were sampled at random from pay records and asked to complete an anonymous job satisfaction survey.                      Degree of Satisfaction  \text { Degree of Satisfaction }
 Pay Type  Satisfied  Neutral  Dissatisfied  Row Total  Salaried 40101060 Hourly 805050180 Col Total 1206060240\begin{array} { |c | c | c | c | c | } \hline{ \text { Pay Type } } & \text { Satisfied } & { \text { Neutral } } & \text { Dissatisfied } & \text { Row Total } \\\hline \text { Salaried } & 40 & 10 & 10 & 60 \\\hline\text { Hourly } & 80 & 50 & 50 & 180 \\\hline \text { Col Total } & 120 & 60 & 60 & 240\\\hline\end{array} For a chi-square test of independence, degrees of freedom would be:

A)2
B)3
C)4
D)6
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65
As an independent project, a team of statistics students tabulated the types of vehicles that were parked in four different suburban shopping malls.                           MrN Location \text { MrN Location }
 Vehicle Type  Somerset  Oakland  Great Lakes  Jamestown  Row Tolal  Car 44493664193Minivan 2115181367 Full-size Van 233210 SUN 1927261284Truck 13617946 Col Total100100100100400\begin{array}{|c|c|c|c|c|c|}\hline\text { Vehicle Type }& \text { Somerset } & \text { Oakland } & \text { Great Lakes } & \text { Jamestown }&\text { Row Tolal }\\\hline \text { Car }&44 & 49 & 36 & 64&193 \\\hline \text {Minivan }&21 & 15 & 18 & 13 &67\\\hline \text { Full-size Van }&2 & 3 & 3 & 2 &10\\\hline \text { SUN }&19 & 27 & 26 & 12&84 \\\hline \text {Truck }&13 & 6 & 17 & 9 &46\\\hline \text { Col Total}&100 & 100 & 100 & 100 &400\\\hline\end{array} Assuming independence, the expected frequency of SUVs in Jamestown is:

A)12.
B)21.
C)75.
D)60.
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66
Employees of OxCo Mfg. were surveyed to evaluate the company's pension plan. The table below displays some of the results of the survey. <strong>Employees of OxCo Mfg. were surveyed to evaluate the company's pension plan. The table below displays some of the results of the survey.   The expected frequency for the shaded cell in the table would be:</strong> A)163. B)158. C)165. D) 160. The expected frequency for the shaded cell in the table would be:

A)163.
B)158.
C)165.
D) 160.
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67
The table below is a tabulation of opinions of employees of Axolotl Corporation, who were sampled at random from pay records and asked to complete an anonymous job satisfaction survey.                      Degree of Satisfaction  \text { Degree of Satisfaction }
 Pay Type  Satisfied  Neutral  Dissatisfied  Row Total  Salaried 40101060 Hourly 805050180 Col Total 1206060240\begin{array} { |c | c | c | c | c | } \hline{ \text { Pay Type } } & \text { Satisfied } & { \text { Neutral } } & \text { Dissatisfied } & \text { Row Total } \\\hline \text { Salaried } & 40 & 10 & 10 & 60 \\\hline\text { Hourly } & 80 & 50 & 50 & 180 \\\hline \text { Col Total } & 120 & 60 & 60 & 240\\\hline\end{array} Assuming independence, the expected frequency of satisfied hourly employees is:

A)80.
B)90.
C)75.
D)60.
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68
An operations analyst counted the number of arrivals per minute at an ATM in each of 30 randomly chosen minutes. The results were: 0, 3, 3, 2, 1, 0, 1, 0, 0, 1, 1, 1, 2, 1, 0, 1, 0, 1, 2, 1, 1, 2, 1, 0, 1, 2, 0, 1, 0, 1. For the Poisson goodness-of-fit test, what is the expected frequency of the data value X = 1?

A)Impossible to determine.
B)11.04
C)1.00
D)2.47
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69
The table below is a tabulation of opinions of employees of Axolotl Corporation, who were sampled at random from pay records and asked to complete an anonymous job satisfaction survey.                      Degree of Satisfaction  \text { Degree of Satisfaction }
 Pay Type  Satisfied  Neutral  Dissatisfied  Row Total  Salaried 40101060 Hourly 805050180 Col Total 1206060240\begin{array} { |c | c | c | c | c | } \hline{ \text { Pay Type } } & \text { Satisfied } & { \text { Neutral } } & \text { Dissatisfied } & \text { Row Total } \\\hline \text { Salaried } & 40 & 10 & 10 & 60 \\\hline\text { Hourly } & 80 & 50 & 50 & 180 \\\hline \text { Col Total } & 120 & 60 & 60 & 240\\\hline\end{array} For a chi-square test of independence, the critical value for ? = .01 is:

A)9.210.
B)4.605.
C)11.34.
D)16.81.
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70
You test a hypothesis of independence of two variables. The number of observations is 500 and you have classified the data into a 4 by 4 contingency table. The test statistic has __________ degrees of freedom.

A)16
B)9
C)499
D)498
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71
Students in an introductory college economics class were asked how many credits they had earned in college, and how certain they were about their choice of major. Their replies are summarized below.                           Degree of Certainty  \text { Degree of Certainty }
 Credits Earned  Very Uncertain  Somewhat Certain  Very Certain Row Tot Under 10 1416646 10 through 591682044 60 or more 2142238 Col Total 423848128\begin{array} { |c | c | c | c | c | } \hline \text { Credits Earned }&{ \text { Very Uncertain } } & \text { Somewhat Certain } & { \text { Very Certain} } & \text { Row Tot } \\\hline \text {Under 10 } & 14 & 16 & 6 &46 \\\hline\text { 10 through 59} & 16& 8 & 20 & 44 \\\hline \text { 60 or more } & 2 & 14 & 22 & 38\\\hline \text { Col Total } & 42 & 38 & 48 & 128\\\hline\end{array} Assuming independence, the expected frequency of very uncertain students with 60 credits or more is:

A)12.47.
B)2.00.
C)14.56.
D)11.09.
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72
You want to test the hypothesis that the prime rate and inflation are independent. The following table is prepared for the test on the basis of the results of a random sample, collected in various countries and various time periods: <strong>You want to test the hypothesis that the prime rate and inflation are independent. The following table is prepared for the test on the basis of the results of a random sample, collected in various countries and various time periods:   Using α = .05, what is the critical value of the test statistic that you would use?</strong> A)3.841 B)12.59 C)5.991 D)7.815 Using α = .05, what is the critical value of the test statistic that you would use?

A)3.841
B)12.59
C)5.991
D)7.815
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73
Students in an introductory college economics class were asked how many credits they had earned in college, and how certain they were about their choice of major. Their replies are summarized below.                           Degree of Certainty  \text { Degree of Certainty }
 Credits Earned  Very Uncertain  Somewhat Certain  Very Certain Row Tot Under 10 1416646 10 through 591682044 60 or more 2142238 Col Total 423848128\begin{array} { |c | c | c | c | c | } \hline \text { Credits Earned }&{ \text { Very Uncertain } } & \text { Somewhat Certain } & { \text { Very Certain} } & \text { Row Tot } \\\hline \text {Under 10 } & 14 & 16 & 6 &46 \\\hline\text { 10 through 59} & 16& 8 & 20 & 44 \\\hline \text { 60 or more } & 2 & 14 & 22 & 38\\\hline \text { Col Total } & 42 & 38 & 48 & 128\\\hline\end{array} For a chi-square test of independence, degrees of freedom would be:

A)2.
B)9.
C)4.
D)127.
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74
Students in an introductory college economics class were asked how many credits they had earned in college, and how certain they were about their choice of major. Their replies are summarized below.                           Degree of Certainty  \text { Degree of Certainty }
 Credits Earned  Very Uncertain  Somewhat Certain  Very Certain Row Tot Under 10 1416646 10 through 591682044 60 or more 2142238 Col Total 423848128\begin{array} { |c | c | c | c | c | } \hline \text { Credits Earned }&{ \text { Very Uncertain } } & \text { Somewhat Certain } & { \text { Very Certain} } & \text { Row Tot } \\\hline \text {Under 10 } & 14 & 16 & 6 &46 \\\hline\text { 10 through 59} & 16& 8 & 20 & 44 \\\hline \text { 60 or more } & 2 & 14 & 22 & 38\\\hline \text { Col Total } & 42 & 38 & 48 & 128\\\hline\end{array} Under the assumption of independence, the expected frequency in the upper left cell is:

A)15.09.
B)24.00.
C)19.72.
D)20.22.
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75
Employees of OxCo Mfg. were surveyed to evaluate the company's pension plan. The table below displays some of the results of the survey. <strong>Employees of OxCo Mfg. were surveyed to evaluate the company's pension plan. The table below displays some of the results of the survey.   The appropriate conclusion would be:</strong> A)do not reject H<sub>0</sub>. B)reject H<sub>0</sub> at α = .10. C)reject H<sub>0</sub> at α = .05. D)reject H<sub>0</sub> at α = .01. The appropriate conclusion would be:

A)do not reject H0.
B)reject H0 at α = .10.
C)reject H0 at α = .05.
D)reject H0 at α = .01.
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76
Employees of OxCo Mfg. were surveyed to evaluate the company's pension plan. The table below displays some of the results of the survey. <strong>Employees of OxCo Mfg. were surveyed to evaluate the company's pension plan. The table below displays some of the results of the survey.   Degrees of freedom for this test (shaded cell below the table) would be:</strong> A)6. B)7. C)799. D)12. Degrees of freedom for this test (shaded cell below the table) would be:

A)6.
B)7.
C)799.
D)12.
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77
You want to test the hypothesis that the prime rate and inflation are independent. The following table is prepared for the test on the basis of the results of a random sample, collected in various countries and various time periods: <strong>You want to test the hypothesis that the prime rate and inflation are independent. The following table is prepared for the test on the basis of the results of a random sample, collected in various countries and various time periods:   The expected frequency for the shaded cell is:</strong> A)22.5. B)30. C)40. D)40.5. The expected frequency for the shaded cell is:

A)22.5.
B)30.
C)40.
D)40.5.
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78
You want to test the hypothesis that the prime rate and inflation are independent. The following table is prepared for the test on the basis of the results of a random sample, collected in various countries and various time periods: <strong>You want to test the hypothesis that the prime rate and inflation are independent. The following table is prepared for the test on the basis of the results of a random sample, collected in various countries and various time periods:   What is the value of the test statistic?</strong> A)306.25 B)0.00 C)54.44 D)13.61 What is the value of the test statistic?

A)306.25
B)0.00
C)54.44
D)13.61
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79
Students in an introductory college economics class were asked how many credits they had earned in college, and how certain they were about their choice of major. Their replies are summarized below.                           Degree of Certainty  \text { Degree of Certainty }
 Credits Earned  Very Uncertain  Somewhat Certain  Very Certain Row Tot Under 10 1416646 10 through 591682044 60 or more 2142238 Col Total 423848128\begin{array} { |c | c | c | c | c | } \hline \text { Credits Earned }&{ \text { Very Uncertain } } & \text { Somewhat Certain } & { \text { Very Certain} } & \text { Row Tot } \\\hline \text {Under 10 } & 14 & 16 & 6 &46 \\\hline\text { 10 through 59} & 16& 8 & 20 & 44 \\\hline \text { 60 or more } & 2 & 14 & 22 & 38\\\hline \text { Col Total } & 42 & 38 & 48 & 128\\\hline\end{array} For a chi-square test of independence, the critical value for ? = .05 is:

A)5.991.
B)7.815.
C)9.488.
D)16.92.
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80
Students in an introductory college economics class were asked how many credits they had earned in college, and how certain they were about their choice of major. Their replies are summarized below.                           Degree of Certainty  \text { Degree of Certainty }
 Credits Earned  Very Uncertain  Somewhat Certain  Very Certain Row Tot Under 10 1416646 10 through 591682044 60 or more 2142238 Col Total 423848128\begin{array} { |c | c | c | c | c | } \hline \text { Credits Earned }&{ \text { Very Uncertain } } & \text { Somewhat Certain } & { \text { Very Certain} } & \text { Row Tot } \\\hline \text {Under 10 } & 14 & 16 & 6 &46 \\\hline\text { 10 through 59} & 16& 8 & 20 & 44 \\\hline \text { 60 or more } & 2 & 14 & 22 & 38\\\hline \text { Col Total } & 42 & 38 & 48 & 128\\\hline\end{array} Which statement is most nearly correct?

A)The contingency table violates Cochran's Rule.
B)Visual inspection of column frequencies suggests independence.
C)At ? = .05 we would easily reject the null hypothesis of independence.
D)At ? = .05 we cannot reject he null hypothesis of independence.
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Unlock Deck
Unlock for access to all 99 flashcards in this deck.