Exam 17: Additional Tests for Nominal Data: Chi-Squared Tests

A chi-squared test for independence with 10 degrees of freedom results in a test statistic of 17.894. Using the chi-squared table, the most accurate statement that can be made about the p-value for this test is that 0.05 < p-value < 0.10.

Like that of the Student t-distribution, the shape of the chi-squared distribution depends on its number of degrees of freedom.

A study of education levels of 500 voters and their political party affiliations in a particular state in Australia showed the following results. Party Affliation Education level completed Liberal Other Labour Primary education 40 20 80 Secondary education 70 30 60 Tertiary education 90 50 60 Is education level independent of political affiliation? Test at the 5% level of significance.

The number of degrees of freedom for a contingency table with r rows and c columns is (r - 1)(c - 1), provided that both r and c follow the rule of five.

A biology professor claimed that the proportions of grades in his classes are the same. A sample of 100 students showed the following frequencies. Grade A B C D F Frequency 14 23 27 26 10 Determine the rejection region at the 5% significance level.

The president of a large university has been studying the relationship between male/female supervisory structures in his institution and the level of employees' job satisfaction. The results of a recent survey are shown in the table below. Conduct a test at the 5% significance level to determine whether the level of job satisfaction depends on the boss/employee gender relationship. Boss/ Employee Level of Satisfaction Male/Female Female/Male Male/Male Female/Female Satisfied 60 15 50 15 Neutral 27 45 48 50 Dissatisfied 13 32 12 55

The null hypothesis states that the sample data came from a normally distributed population. The researcher calculates the sample mean and the sample standard deviation from the data. The data arrangement consisted of seven categories. Using a 0.05 significance level, the appropriate critical value for this chi-squared test for normality is 11.0705.

A major insurance firm interviewed a random sample of 1500 college students to find out the type of life insurance preferred, if any. The results follow: Insurance Preference Gender Term Whole life No insurance Female 170 110 470 Male 195 75 480 Is there evidence that the life insurance preference of male students is different to that of female students? Test using the 5% level of significance.

The following data are believed to have come from a normal probability distribution. 26 21 25 20 21 29 26 23 22 24 24 30 23 32 26 24 32 16 36 26 21 31 26 23 32 35 40 30 14 26 46 27 33 25 27 21 26 18 29 36 The mean of this sample equals 26.80, and the standard deviation equals 6.378. Use the goodness-of-fit test at the 5% significance level to test this claim.

Which of the following uses the Chi-square distribution?

Which of the following best describes the rejection region in a chi-squared test of independence?

Whenever the expected frequency of a cell is less than 5, one remedy for this condition is to increase the size of the sample.

A chi-squared test for independence with 6 degrees of freedom results in a test statistic of 13.25. Using the chi-squared table, the most accurate statement that can be made about the p-value for this test is that 0.025 < p-value < 0.05.

A simple random sample of 150 people was taken to investigate whether there is a link between smoking status and having had a tertiary education. Test at the 5% level of significance. Non-smoker Smoker None-tertiary educated 42 28 Terti ary educated 23 57

To determine whether a single coin is fair, the coin was tossed 200 times. The observed frequencies with which each of the two sides of the coin turned up are recorded as 112 heads and 88 tails. Is there sufficient evidence at the 5% significance level to allow you to conclude that the coin is not fair?

A caterer proposes to serve four main courses. For planning purposes, the caterer expects that the proportions of each that will be selected by customers will be: Selection Proportion Roast beef 0.50 Chicken 0.20 Fish 0.10 Vegeterian 0.20 Of the first 100 customers, 44 select roast beef, 24 select chicken, 13 select fish, and 19 select the vegetarian meal. Should the caterer revise the estimates? Use $\alpha$ = 0.05.

Consider a multinomial experiment involving n = 200 trials and k = 5 cells. The observed frequencies resulting from the experiment are shown in the following table: Cell 1 2 3 4 5 Frequency 16 44 56 48 36 The null hypothesis to be tested is as follows. H₀ : p₁ = 0.05, p₂ = 0.25, p₃ = 0.35, p₄ = 0.20, p₅ = 0.15. Test the hypothesis at the 5% level of significance.

Which of the following statements is not correct?

If we want to test for differences between two populations of nominal data with exactly two categories, we can employ either the z-test of $p _ { 1 } - p _ { 2 }$ , or the chi-squared test of a contingency table. (Squaring the value of the z-statistic yields the value of the $x ^ { 2 }$ -statistic.)

A test for independence is applied to a contingency table with 4 rows and 4 columns for two nominal variables. The number of degrees of freedom for this test will be 9.