Statistical Inference: Chapter 11.1-2 - Hypothesis Testing and Confidence Intervals for Comparing Two Population Means

1. Statistical Inference: Chapter 11

2. 1. 2. Chapter Goals When you have completed this chapter, you will be able to: Explain the the difference between dependentand independent samples. Conduct a test of hypothesis and obtain a confidence interval estimatefor the difference between … … two population means using independent samples. … two population means using matched pair sample. … two population proportions.

3. Hypothesis Testing Recall State the null and alternate hypotheses Step 1 Select the level of significance Step 2 Identify the test statistic Step 3 State the decision rule Step 4 Step 5 Compute the value of the test statistic and make a decision Do not reject H0 Reject H0 and acceptH1

4. Comparing Two Populations We wish to know whether the the distributionof thedifferences in sample meanshas a mean of 0. If both samples contain at least 30 observations we use the z distribution as the test statistic.

5. shape No assumptions about the of the populations are required. X X - z = 2 1 s s 2 2 + 2 1 n1 n2 Comparing Two Populations The samples are from independent populations. The formula for computing the value of zis:

6. Q uestion Solve Two colleges are located in York Region. The local paper recently reported that the meanstarting salary of a graduate from College 1 is $38,000 with a standard deviation of $6,000 for a sample of 40 graduates. The same article reported the mean starting salary of a graduate from College 2 is $35,000 with a standard deviation of $7,000 for a sample of 35 graduates. At the .01 significance level can we conclude the mean salary of College 1 is more?

7. Step 1 State the null and alternate hypotheses Step 2 Select the level of significance Step 3 Identify the test statistic Step 4 State the decision rule - X X 1 2 = z 2 2 s s 1 2 + n n Step 5 Compute the test statistic and make a decision 1 2 $38,000 - $35,000 z = 2 2 ($ 6 , 000) ($ 7 , 000 ) + 40 35 Hypothesis Test H0: µ1 = µ2 H1: µ1 > µ2  = 0.01 Because both samples are more than 30, the test statistic is Z Reject H0 if z > 2.33 = 1.98 Conclusion: Do not reject the null hypothesis; insufficient evidence.

8. Look up in Table Step 4 State the decision rule The p-value is: alternate .5000 - p(z> 1.98) = .4761 = .0239 Since pof .0239 is greater than the chosen alpha of 0.01, there is insufficient evidence to reject H0 0 1.98

9. Small Sample Tests of Means If one or more of thesamples contain less than 30 observations we use the t distribution as the test statistic. Required Assumptions: Both populations must follow the normal distribution. The populations must have equal standard deviations. The samples are from independent populations.

10. - + - ( n 1 ) s ( n 1 ) s 2 2 = 2 s 1 1 2 2 p n n + - 2 1 2 - X X 1 2 = T æ ö 1 1 + 2 ç ÷ s ç ÷ p n n è ø 1 2 Small Sample Tests of Means Finding the value of the test statistic Two steps needed: • Pool the sample variances • Determine the value of T from the formula

11. Q uestion Solve A recent EPA study compared the highway fuel economy of domestic and imported passenger cars. A sample of15 domestic cars revealed ameanof33.7 mpg with a standard deviation of 2.4 mpg. A sample of 12 imported cars revealed a mean of 35.7 mpg with a standard deviation of 3.9. At the .05 significance level can the EPA conclude that the mpg is higher on the imported cars?

12. Step 1 State the null and alternate hypotheses Step 2 Select the level of significance Step 3 Identify the test statistic - + - ( n 1 ) s ( n 1 ) s 2 2 = 2 s 1 1 2 2 p n n Step 4 + - State the decision rule 2 1 2 Step 5 Compute the test statistic and make a decision Solution continues… Hypothesis Test H0: µd = µ1 H1: µd < µ1  = 0.05 Because both samples are less than 30, the test statistic is t Reject H0 if t < 1.708 There are 25 d.f. Pooled Variances

13. - + - ( n 1 ) s ( n 1 ) s 2 2 = 2 s 1 1 2 2 p n n + - 2 - + - 2 2 s 2 ( 15 1 )( 2 . 4 ) ( 12 1 )( 3 . 9 ) = 1 2 p = 9 . 918 + - 15 12 2 - - T 33 . 7 35 . 7 = Step 5 Compute the test statistic and make a decision 1 1 1 1 + + 9 . 918 2 s X X p 15 12 1 2 n n 1 2 Hypothesis Test Pooled Variances ComputeT = Conclusion: H0 cannot be rejected… insufficient evidence.

14. Hypothesis Testing Involving Paired Observations Independent samples are samples that are not related in any way Dependent samples are samples that are paired or related in same fashion Examples If you wished to measure the effectiveness of a new diet, you would weigh at the start and at the finish of the programme! If you wished to buy a car, you would look at the same car at two (or more) different dealerships and compare the prices.

15. D = T s / n D D Hypothesis Testing Involving Paired Observations Use the following testwhen the samples aredependent: where …is the mean of the differences sD …is the standard deviation of the differences n …is the number of pairs (differences)

16. At the .05 significance level, can the testing agency conclude that there is a difference in the rental charged? An independent testing agency is comparing the daily rental cost for renting a compact car from Hertz and Avis. A random sample of eight cities revealed the following information:

17. Step 1 H1: µD=01 State the null and alternate hypotheses Step 2 Select the level of significance Step 3 Identify the test statistic Step 4 State the decision rule Step 5 Compute the test statistic and make a decision D = T s / n D Solution continues… Hypothesis Test H0: µD = 0  = 0.05 Because both samples are less than 30, the test distribution is t Reject H0 if T < -2.365or T > 2.365 There are 7 d.f.

18. 4 2 0 0 5 25 Step 5 Compute the test statistic and make a decision -3 9 D = T 2 4 s / n -4 16 D Solution continues… Hypothesis Test 2 4 16 4

19. Step 5 Compute the test statistic and make a decision D S 8 . 0 = D = = 1 . 00 n 8 s = D 2 2 S S D D - 82 - n 8 - 8 4 1 2 0 0 1 . 00 5 25 3 . 1623 8 D -3 9 = T 2 4 s / n 8 78 - D n 1 -4 16 Hypothesis Test 2 4 16 4 78 = = 3.1623 = T = 0.894 Because 0.894 < 2.365, we can not reject H0.There isno significant difference in the prices Conclusion:

20. + X X 1 2 = ˆ p + n n 1 2 z - ˆ ˆ p p = 1 2 é ù 1 - ˆ ˆ p ( 1 p ) ê ú + n n ë û 1 2 Two Sample Tests of Proportions We investigate whether two samples came from populations with an equal proportion of success. The two samples are pooled using the following formula: where X1 andX2 refer to the number of successes in the respective samplesn1and n2 Test Statistic

21. Q uestion Are unmarried workers more likely to be absent from work than marriedworkers? A sample of 250 married workers showed 22 missed more than 5 days last year, while a sample of 300 unmarried workers showed 35 missed more thanfive days. Use a .05 significance level.

22. H0: pupm = Step 1 State the null and alternate hypotheses Step 2 Select the level of significance Step 3 Identify the test statistic Step 4 State the decision rule Step 5 Compute the test statistic and make a decision + X X 1 2 = ˆ p + n n 1 2 Solution continues… Hypothesis Test H1: Pu>Pm  = 0.05 Because both samples are large the z distribution is used. Reject H0 if z > 1.645

23. = .1036 35 22 - z = 300 250 - ˆ ˆ p p 1 é ù 2 1 - - ) .1036 ( 1 . 1036 . 1036 ( 1 . 1036 ) + ú ê + n n ë û Step 5 300 250 Compute the test statistic and make a decision 2 1 + X X - ˆ ˆ p ( 1 p ) 1 2 = ˆ p + n n 1 2 Conclusion: Hypothesis Test 22 + 35 = 250 + 300 = = 1.10

24. Conclusion: The null hypothesis is not rejected. We cannot conclude that a higher proportion of unmarried workers miss more days in a year than the marriedworkers. Hypothesis Test The p-value is: P(z > 1.10) = .5000 - .3643 = .1457

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26. This completes Chapter 11