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MATH221 Integrated Learning System Power Point Lecture Week 5 Hypothesis Testing Hypothesis Testing for the Mean: Small Samples The only difference between hypothesis testing for the population mean with small (n < 30) samples and hypothesis testing for the mean with large (n 30) samples, which we covered on-site, is the test statistic used in the test. For small samples we will use the t statistic. The t-test for a Mean x t s/ n Example A computer salesman believes that the mean repair cost for damaged computers is more than $95. To test this claim, you determine the repair cost for seven randomly selected computers and find that the mean repair cost is $100 per computer with a standard deviation of $42.50. At = 0.01, do you have enough evidence to support the repairman’s claim? Solution First we state the Hypothesis. Ho: $95 Ha: > $95 (claim) n = 7, so the degrees of freedom are 6. Solution (continued) This is a right-tailed test so looking up the critical value for t with = 0.01 and d.f. = 6 we get t0.01,6 = 3.143. So we will reject Ho if t > 3.143. t 100 95 42.5 / 7 0.311 So we fail to reject Ho and conclude that there is not enough evidence to support the repairman’s claim. Using Technology: TI 83 T-Test Inpt:Data Stats o:95 x:100 Sx:42.5 n:7 :o <o >o Calculate Draw Using Technology: TI 83 (Cont.) T-Test >95 t=.3112648601 p=.3830651297 x=100 Sx=42.5 n=7 Student Exercise An environmentalist estimates that the mean waste recycled by American adults is more than one pound per person per day. You want to test this claim. You find that the mean waste recycled per person per day for a random sample of 12 American adults is 1.2 pounds and the standard deviation is 0.3 pounds. At = 0.05, can you support this claim? Hypothesis Testing for Proportions Earlier we learned that if np > 5 and nq > 5 then the sampling distribution for the sample proportions is normal with: pˆ p and pˆ pq / n The test statistic for the hypothesis test of proportions is: pˆ p z pq / n Example A medical researcher estimates that no more than 55% of American adults eat breakfast everyday. In a random sample of 250 adults, 56.4% say that they eat breakfast everyday. At = 0.01, is their enough evidence to reject the researcher’s claim? Solution Ho: p 0.5 (claim) Ha: p > 0.5 Right tailed test so z0.01 = 2.326 Reject Ho if z > 2.326 z .564 .55 .55(.45) / 250 0.445 So we fail to reject Ho and conclude that there is not enough evidence to reject the researcher’s claim. Student Exercise In your work for a business regulatory agency you find that in a sample of 1762 Americans, 1004 of them believe that government regulation of business does more harm than good. At = 0.02, can you reject the claim that 60% of Americans have this view? Hypothesis Testing for Variance and Standard Dev. If we want to manufacture items, say golf balls, that have to be nearly exactly alike in all respects, we must be concerned with controlling variation. If the population is normal, you can test the variance and standard deviation of the population using the chi-squared test with n – 1 degrees of freedom. Guidelines for Chi-Square Test 1. Write the null and alternative hypotheses. 2. Specify . 3. The Critical values for the 2 distribution are found in Table 6 of Appendix B in your text. 1. For a right-tailed test use the value corresponding to and the d.f. 2. For a left tailed test use the value corresponding to 1 - and the d.f. 3. For a two-tailed test use the values corresponding to the d.f. and /2, and 1 - /2. 2 Test Statistic 2 n 1s 2 2 Example A travel agency estimates that the standard deviation of room rates in a certain city is no more than $25. You work for a consumer advocacy group and are asked to test this claim. You find that a random sample of 13 hotel room rates has a standard deviation of $27.50. At = 0.01, do you have enough evidence to reject the agency’s claim? Solution Ho: $25.00 (claim) Ha: > $25.00 d.f. = 12 This is a right-tailed test so we look up 2 for = 0.01 and d.f. = 12. The critical value is 2 = 26.217 Reject Ho if 2 > 26.217 Solution (continued) Test Statistic: 2 13 1 27 . 7 2 2 25 14.732 So we fail to reject the null hypothesis and conclude that there is not enough evidence to reject the claim. Student Exercise A hospital spokesperson claims that the standard deviation of the waiting times experienced by patients in its minor emergency department is no more than 0.5 minutes. You doubt the validity of this claim. If a random sample of 25 waiting times has a standard deviation of 0.7 minutes, can you reject the spokesperson’s claim? Use an of 0.10.