Example 11-3, pg. 495 The Chi-Square Goodness-of-Fit Test

Transcription

1 132 Chapter 11 Chi-Square Tests Chi-Square Tests Chapter 11 Section 11.2 Example 11-3, pg. 495 The Chi-Square Goodness-of-Fit Test A bank manager wanted to investigate if the percentage of people who use the ATM machine is the same for each weekday. The data, found on page 495 of the textbook, is shown below. Day Monday Tuesday Wednesday Thursday Friday Number of Users Enter the data into the Data Window. Put the Days of the Week into C1 and name the column Day. Enter the Number of Users into C2 and name it Observed. Enter the proportions from the null hypothesis into C3 and name it Distribution. These five proportions should all be equal to 0.20.

2 Section To calculate the expected frequencies, multiply the proportions times the sample size, which is 1200 in this example. Click on Calc Calculator. You will Store the result in C4, and calculate the Expression C3*1200. Click on OK. Name C4 Expected since it now contains the expected frequencies.

3 134 Chapter 11 Chi-Square Tests Next, calculate the chi-square test statistic, (O - E) 2 / E. Click on Calc Calculator. You will Store the result in C5, and calculate the Expression (C2 - C4)**2 / C4.

4 Section Click on OK and C5 should contain the calculated values. Now, just add up the values in C5 and the sum is the test statistic. Click on Calc Column Statistics. Select Sum and enter C5 for the Input Variable.

5 136 Chapter 11 Chi-Square Tests Click on OK and the Chi-Square test statistic will be displayed in the Session Window. In this example, the test statistic is Next, calculate the P-value to help you decide if you should reject the null hypothesis. Click on Calc Probability Distributions Chi-square. Select Cumulative Probability with Noncentrality parameter set at 0 and enter 4 Degrees of Freedom. Enter the value of the test statistic, , for the Input Constant. Click on OK and P(X ) will be in the Session Window.

6 Section The P(X ) is The P-value is P(X ), which is 1 - P(X ). This value is = Since this P-value is smaller than α =.05, you should reject the null hypothesis. The proportion of people using the ATM machine is not the same for each day of the week.

7 138 Chapter 11 Chi-Square Tests Section 11.4 Example 11-6, pg. 505 Chi-Square Independence Test A random sample of 300 adults was selected and they were asked if they favor giving more freedom to schoolteachers to punish students for violence and lack of discipline. The data, found on page 503, of the textbook is shown below. In Favor (F) Against (A) No Opinion (N) Men (M) Women (W) Does the sample provide sufficient evidence to conclude that gender and opinions of adults are dependent? To test this hypothesis, enter the data into the MINITAB Data Window. First label the columns: use Gender for C1, use In Favor (F) for C2, use Against (A) for C3, and use No Opinion (N) for C4. Now enter the data into the appropriate columns. Do not enter any totals. To perform the chi-square independence test, click on Stat Tables Chisquare Test. On the input screen, select C2 C4 for the Columns containing the table.

8 Section Click on OK and the test results will be displayed in the Session Window.

9 140 Chapter 11 Chi-Square Tests Notice that the Minitab output contains the expected frequencies below each observed frequency in the table. The Chi-Square Contributions are printed below the expected frequencies. The test statistic is Chi-Sq = and the P-value =.016. Since this P-value is larger than α =.01, you should not reject the null hypothesis. Thus, there is not enough evidence to conclude that gender and opinion are dependent.

10 Section Example 11-8, pg. 509 Test of Homogeneity Consider the data on income distribution for households in California and Wisconsin shown below, and found on page 508 of the textbook. California Wisconsin High Income Medium Income Low Income Enter the data into the Data Window. We wish to test, at α =.025, the null hypothesis that the distribution of households with regard to income levels is similar (homogeneous) for the two states. To perform the chi-square test, click on Stat Tables Chi-square Test. On the input screen, select C2 C3 for the Columns containing the table. Click on OK and the test results will be displayed in the Session Window.

11 142 Chapter 11 Chi-Square Tests Notice that the test statistic is Chi-Sq = and the P-value =.114. Since this P-value is larger than α =.025, you should NOT reject the null hypothesis. Thus, we conclude that the distribution of households with regard to income appears to be similar (homogeneous) in California and Wisconsin. Notice that the Minitab output contains the expected frequencies below each observed frequency in the table. The Chi-Square Contributions are printed below the expected frequencies.

12 Section Section 11.5 Example 11-10, pg. 517 Hypothesis Tests About the Population Variance MINITAB does not have a hypothesis test for the population variance, however, if you first calculate the test statistic, MINITAB can calculate the p-value for you. You can use the p-value to help you decide whether or not to reject the null hypothesis. A machine that fills packages of Cocoa Cookies is set up so that the average net weight of a package should be 32 ounces with a variance of.015 square ounces. A sample of 25 packages from the production line gave a sample variance of.029. At α=.01, test if the filling machine needs an adjustment. Calculate the test statistic: χ 2 = (25-1)*(.029)/(.015) = Once you have calculated the χ 2 - test statistic, MINITAB can calculate the exact P-value of the test. Click on Calc Probability distributions Chi-square. Select Cumulative Probability and enter 24 Degrees of Freedom. Next, click on Input Constant and enter the test statistic, Click on OK and the output will be displayed in the Session Window. Cumulative Distribution Function Chi-Square with 24 DF x P( X <= x ) Since this is an upper-tailed test (H 0 : σ 2 =.015 vs. H a :σ 2 >.015), you should subtract the p-value from 1. Thus, the P-value = =.004. Since this value is smaller than α =.01, you reject the null hypothesis. The machine should be stopped and adjusted.

13 144 Chapter 11 Chi-Square Tests Suggested Exercises Section 11.2 pp : 11.14, 11.15, Section 11.4 pp : 11.29, 11.30, 11.31, 11.38, Section 11.5 pp. 520: 11.46b, 11.47b Technology Assignments pp : TA11.1, TA11.2, TA11.3