1. 1 Claims about two large means E.g., “South students have a higher average SAT score that Monmouth Regional Students” The two samples are independent. The samples from one population are not related to the other samples Both samples are simple random samples. Large and small samples Same general procedure –Test statistic requires two samples from each population –Classic method –P-value (calculator)

2. 2 Hypothesis Testing: Claims 1.State the claim: South students have the same mean SAT scores as North students μ 1 = μ 2 2.State the opposite claim: μ 1  μ 2 3.Pick your hypothesis 4.Sample: XSn South171020060 North1680280110

3. 3 Hypothesis Tests: Test Statistic 5.Test statistic –(μ 1 – μ 2 ) will almost always be 0 6.Critical value: –Z table if both samples are 30 or more, –T table otherwise

4. Testing two sample means Original claimPick null and alternate hypothesis Opposite claim Degree of confidence First sampleX-bar =s =n = Second sampleX-bar =s =n = Test statistic Two tailed (=) Right tailed () Left tailed Critical value P-value Conclusion 4

5. 5 Using the Calculator Finding the p-value STAT→TEST→2-SampleZTest STAT→TEST→2-SampleTTest Enter the mean, standard deviation, and sample size for both samples Select H 1 Select Calculate If p-value is less than alpha, reject.

6. Testing two sample means Original claimPick null and alternate hypothesis Opposite claim Degree of confidence First sampleX-bar =s =n = Second sampleX-bar =s =n = Test statistic Two tailed (=) Right tailed () Left tailed Critical value P-value Conclusion 6

7. 7 Live example/Your turn It rains more on Saturday than Monday Books are more expensive at UMass than DCC Xsn Monday0.10.2625 Saturday0.140.2942 Xsn UMass65.1228.0340 DCC57.6214.6835

8. Testing two sample means Original claimPick null and alternate hypothesis Opposite claim Degree of confidence First sampleX-bar =s =n = Second sampleX-bar =s =n = Test statistic Two tailed (=) Right tailed () Left tailed Critical value P-value Conclusion 8

9. Testing two sample means Original claimPick null and alternate hypothesis Opposite claim Degree of confidence First sampleX-bar =s =n = Second sampleX-bar =s =n = Test statistic Two tailed (=) Right tailed () Left tailed Critical value P-value Conclusion 9

10. Homework 1.Use a 0.05 significance level to test the claim that the two populations have the same mean: –Treatment group: n = 50, x-bar = 7.00, s = 1.00 –Placebo group:n = 100, x-bar = 6.00, s = 2.00 2.Use a 90% degree of confidence to test the claim that the two populations have different means: –Math majors: n = 40, x-bar = 75.0, s = 15.0 –English majors:n = 60, x-bar = 70.0, s = 14.00 10

11. More Homework 3.A sample of 290 children not wearing seatbelts during car accidents resulted in an average of 1.39 days in the ICU (s = 3.06). On the other hand, 123 children wearing seat belts averaged 0.83 days in the ICU (s = 1.77). Test the claim that wearing seatbelts results in less severe injuries during car accidents. 4.Test the claim that American cars are as reliable as Japanese cars. 35 Japanese cars had an average repair bill of $650 with a standard deviation of $250. 41 American cards has a mean repair bill of $720 with a standard deviation of $390. 11

12. Still More Homework 5.A recent poll at a local college showed that the average age of 217 students’ cars was 7.89. (s = 3.67). The average age of 152 faculty cars was 5.99 (s = 3.65). Using a 0.05 significance level to test the claim that student cars are older than faculty cars. 6.Thirty five pennies minted before 1983 weighed an average of 3.0748 grams (s = 0.0391). Thirty five pennies minted after 1983 weighed an average of 2.5002 grams (s = 0.0149). Test the claim that the weight of pennies did not change after 1983. 12

13. Homework #1 Original claimμ1 = μ2 H 0 Pick null and alternate hypothesis Opposite claimμ1 ≠ μ2 H 1 Degree of confidence95% First sampleX-bar = 7.00s = 1.00n = 50 Second sampleX-bar = 6.00s = 2.00n = 100 Test statisticz = 4.08 Two tailed (=) Right tailed () Left tailed Critical value P-value0.000044 < 0.05 ConclusionReject null, reject original claim 13

14. Homework #2 Original claimμ1 ≠ μ2 H 1 Pick null and alternate hypothesis Opposite claimμ1 = μ2 H 0 Degree of confidence90% First sampleX-bar =s =n = Second sampleX-bar =s =n = Test statistic Two tailed (=) Right tailed () Left tailed Critical value P-value Conclusion 14

15. Homework #3 Original claimμ1 < μ2 H 1 Pick null and alternate hypothesis Opposite claimμ1 ≥ μ2 H 0 Degree of confidence95% First sample (seat belts)X-bar =s =n = Second sample (not)X-bar =s =n = Test statistic Two tailed (=) Right tailed () Left tailed Critical value P-value Conclusion 15

16. Homework #4 Original claimμ1 = μ2 H 0 Pick null and alternate hypothesis Opposite claimμ1 ≠ μ2 H 1 Degree of confidence95% First sample (USA)X-bar =s =n = Second sample (Japan)X-bar =s =n = Test statistic Two tailed (=) Right tailed () Left tailed Critical value P-value Conclusion 16

17. Homework #5 Original claimμ1 > μ2 H 1 Pick null and alternate hypothesis Opposite claimμ1 ≤ μ2 H 0 Degree of confidence95% First sample (students)X-bar =s =n = Second sample (faculty)X-bar =s =n = Test statistic Two tailed (=) Right tailed () Left tailed Critical value P-value Conclusion 17

18. Homework #6 Original claimμ1 = μ2 H 0 Pick null and alternate hypothesis Opposite claimμ1 ≠ μ2 H 1 Degree of confidence95% First sampleX-bar =s =n = Second sampleX-bar =s =n = Test statistic Two tailed (=) Right tailed () Left tailed Critical value P-value Conclusion 18