Exam 15: Chi-Square Tests

In a goodness-of-fit test, a linear probability plot suggests that the null hypothesis should be rejected.

ECDF tests have an advantage over the chi-square goodness-of-fit test on frequencies because an ECDF test treats observations individually.

The null hypothesis for a chi-square contingency test of independence for two variables always assumes that the variables are independent.

As an independent project, a team of statistics students tabulated the types of vehicles that were parked in four different suburban shopping malls.                          $\text { MrN Location }$ Vehicle Type Somerset Oakland Great Lakes Jamestown Row Tolal Car 44 49 36 64 193 Minivan 21 15 18 13 67 Full-size Van 2 3 3 2 10 SUN 19 27 26 12 84 Truck 13 6 17 9 46 Col Total 100 100 100 100 400 For a chi-square test of independence, the critical value for ? = .10 is:

In a chi-square goodness-of-fit test, we lose one degree of freedom for each parameter estimated.

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.

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.

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.                          $\text { Degree of Certainty }$ Credits Earned Very Uncertain Somewhat Certain Very Certain Row Tot Under 10 14 16 6 46 10 through 59 16 8 20 44 60 or more 2 14 22 38 Col Total 42 38 48 128 Assuming independence, the expected frequency of very uncertain students with 60 credits or more is:

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.

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:

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?

An attraction of the Kolmogorov-Smirnov test is that it is fairly easy to do without a computer.

The chi-square test can only be used to assess independence between two variables.

A contingency table shows:

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.                     $\text { Degree of Satisfaction }$ Pay Type Satisfied Neutral Dissatisfied Row Total Salaried 40 10 10 60 Hourly 80 50 50 180 Col Total 120 60 60 240 For a chi-square test of independence, the critical value for ? = .01 is:

Cochran's Rule requires observed frequencies of 5 or more in each cell of a contingency table.

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.

In a chi-square goodness-of-fit test, a sample of n observations has n - 1 degrees of freedom.

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?

The Oxnard Retailers Anti-Theft Alliance (ORATA) published a study that claimed the causes of disappearance of inventory in retail stores were 30 percent shoplifting, 50 percent employee theft, and 20 percent faulty paperwork. The manager of the Melodic Kortholt Outlet performed an audit of the disappearance of 80 items and found the frequencies shown below. She would like to know if her store's experience follows the same pattern as other retailers. Reason Shoplifting Employee Theft Poor Paperwork Frequency 32 38 10 Under the null hypothesis that her store follows the published pattern, the expected number of items that disappeared due to shoplifting is: