Exam 11: Chi-Square and Analysis of Variance

Given below are the analysis of variance results from a Minitab display. Assume that you want to use a 0.05 significance level in testing the null hypothesis that the different samples come from populations with the same mean. -What can you conclude about the equality of the population means? Source DF SS MS F p Factor 3 30 10.00 1.6 0.264 Error 8 50 6.25 Total 11 80

Test the claim that the samples come from populations with the same mean. Assume that the populations are normally distributed with the same variance. -A consumer magazine wants to compare the lifetimes of ballpoint pens of three different types. The magazine takes a random sample of pens of each type in the following table. Brand A Brand B Brand C 260 181 238 218 240 257 184 162 241 219 218 213 Do the data indicate that there is a difference in mean lifetime for the three brands of ballpoint pens? $\text { Use } \alpha = 0.01 \text {. }$

A survey conducted in a small business yielded the results shown in the table. Men Women Health insurance 50 20 No health insurance 30 10 i) Test the claim that health care coverage is independent of gender. Use a 0.05 significance level. ii) Using Yates' correction, replace $\sum { \frac { ( \mathrm { O } - \mathrm { E } ) ^ { 2 } } { E } } \text { with } \sum { \frac { ( | \mathrm { O } - \mathrm { E } | - 0.5 ) ^ { 2 } } { E } }$ and repeat the test. What effect does Yates' correction have on the value of the test statistic?

An observed frequency distribution is as follows: Number of successes 0 1 2 Frequency 41 93 66 i) Assuming a binomial distribution with $\mathrm { n } = 2$ and $\mathrm { p } = 1 / 2$ , use the binomial formula to find the probability corresponding to each category of the table. ii) Using the probabilities found in part (i), find the expected frequency for each category. iii) Use a $0.05$ level of significance to test the claim that the observed frequencies fit a binomial distribution for which $\mathrm { n } = 2$ and $\mathrm { p } = 1 / 2$ .

The test statistic $\text { one-way } \mathrm { ANOVA } \text { is } \mathrm { F } = \frac { \text { variance between samples } } { \text { variance within samples } }$ . Describe variance within samples and variance between samples. What relationship between variance within samples and variance between samples would result in the conclusion that the value of F is significant?

Use the data given below to verify that the t test for independent samples and the ANOVA method are equivalent. 10 19 29 18 11 27 19 30 15 18 16 21 i) Use a t test with a 0.05 significance level to test the claim that the two samples come from populations with the same means. ii) Use the ANOVA method with a 0.05 significance level to test the same claim. iii) Verify that the squares of the t test statistic and the critical value are equal to the F test statistic and critical value.

Use a $\chi ^ { 2 }$ test to test the claim that in the given contingency table, the row variable and the column variable are independent. -Use the sample data below to test whether car color affects the likelihood of being in an accident. Use a significance level of 0.01. Red Blue White Car has been in accident 28 33 36 Car has not been in accident 23 22 30

When using statistical software packages, the critical value is typically not given. What method is used to determine whether you reject or fail to reject the null hypothesis?

Among the four northwestern states, Washington has 51% of the total population, Oregon has 30%, Idaho has 11%, and Montana has 8%. A market researcher selects a sample of 1000 subjects, with 450 in Washington, 340 in Oregon, 150 in Idaho, and 60 in Montana. At the 0.05 significance level, test the claim that the sample of 1000 subjects has a distribution that agrees with the distribution of state populations.

Six independent samples of 100 values each are randomly drawn from populations that are normally distributed with equal variances. You wish to test the claim that $\mu _ { 1 } = \mu _ { 2 } = \mu _ { 3 } =$ $\mu _ { 4 } = \mu _ { 5 } = \mu _ { 6 }$ i) If you test the individual claims $\mu _ { 1 } = \mu _ { 2 } , \mu _ { 1 } = \mu _ { 3 } , \mu _ { 1 } = \mu _ { 4 } , \ldots , \mu _ { 5 } = \mu _ { 6 }$ , how many ways can you pair off the 6 means? ii) Assume that the tests are independent and that for each test of equality between two means, there is a $0.90$ probability of not making a type I error. If all possible pairs of means are tested for equality, what is the probability of making no type I errors? iii) If you use analysis of variance to test the claim that $\mu _ { 1 } = \mu _ { 2 } = \mu _ { 3 } = \mu _ { 4 } = \mu _ { 5 } = \mu _ { 6 }$ at the $0.10$ level of significance, what is the probability of not making a type I error?

Find the critical value. Source DF SS MS F p Factor 3 30 10.00 1.6 0.264 Error 8 50 6.25 Total 11 80

Describe the test of homogeneity. What characteristic distinguishes a test of homogeneity from a test of independence?

List the assumptions for testing hypotheses that three or more means are equivalent.

Define categorical data and give an example.

Test the claim that the samples come from populations with the same mean. Assume that the populations are normally distributed with the same variance. -At the 0.025 significance level, test the claim that the four brands have the same mean if the following sample results have been obtained. Brand A Brand Brand Brand D 15 20 21 15 25 17 22 15 21 22 20 14 23 23 19 23 22 18 22 20 28 28

Use a X² test to test the claim that in the given contingency table, the row variable and the column variable are independent. -The table below shows the age and favorite type of music of 668 randomly selected people. Rock Pop Classical 15-25 50 85 73 25-35 68 91 60 35-45 90 74 77 Use a 5 percent level of significance to test the null hypothesis that age and preferred music type are independent.

Given below are the analysis of variance results from a Minitab display. Assume that you want to use a 0.05 significance level in testing the null hypothesis that the different samples come from populations with the same mean. -Identify the p-value. Source DF SS MS F p Factor 3 13.500 4.500 5.17 0.011 Error 16 13.925 0.870 Total 19 27.425

According to Benford's Law, a variety of different data sets include numbers with leading (first) digits that follow the distribution shown in the table below. Test for goodness-of-fit with Benford's Law. Leading Digit 1 2 3 4 5 6 7 8 9 Benford's law: distribution of leading digits 30.1\% 17.6\% 12.5\% 9.7\% 7.9\% 6.7\% 5.8\% 5.1\% 4.6\% -When working for the Brooklyn District Attorney, investigator Robert Burton analyzed the leading digits of the amounts from 784 checks issued by seven suspect companies. The frequencies were found to be 0, 18, 0, 79, 476, 180, 8, 23, and 0, and those digits correspond to the leading digits of 1, 2, 3, 4, 5, 6, 7, 8, and 9, respectively. If the observed frequencies are substantially different from the frequencies expected with Benford's Law, the check amounts appear to result from fraud. Use a 0.05 significance level to test for goodness-of-fit with Benford's Law. Does it appear that the checks are the result of fraud?

Use a $\chi ^ { 2 }$ test to test the claim that in the given contingency table, the row variable and the column variable are independent. -Tests for adverse reactions to a new drug yielded the results given in the table. At the 0.05 significance level, test the claim that the treatment (drug or placebo) is independent of the reaction (whether or not headaches were experienced). Drug Placebo Headaches 11 7 No headaches 73 91

Find the critical value. Source DF SS MS F p Factor 3 13.500 4.500 5.17 0.011 Error 16 13.925 0.870 Total 19 27.425