1. 1 Objective Compare of two matched-paired means using two samples from each population. Hypothesis Tests and Confidence Intervals of two dependent means use the t-distribution Section 9.4 Inferences About Two Means (Matched Pairs)

2. 2 Definition Two samples are dependent if there is some relationship between the two samples so that each value in one sample is paired with a corresponding value in the other sample. Two samples can be treated as the matched pairs of values.

3. 3 Examples Blood pressure of patients before they are given medicine and after they take it. Predicted temperature (by Weather Forecast) and the actual temperature. Heights of selected people in the morning and their heights by night time. Test scores of selected students in Calculus-I and their scores in Calculus-II.

4. 4 Example 1 First sample: weights of 5 students in April Second sample: their weights in September These weights make 5 matched pairs Third line: differences between April weights and September weights (net change in weight for each student, separately) In our calculations we only use differences (d), not the values in the two samples.

5. 5 Notation d Individual difference between two matched paired values μ d Population mean for the difference of the two values. n Number of paired values in sample d Mean value of the differences in sample s d Standard deviation of differences in sample

6. 6 (1) The sample data are dependent (i.e. they make matched pairs) (2) Either or both the following holds: The number of matched pairs is large (n>30) or The differences have a normal distribution Requirements All requirements must be satisfied to make a Hypothesis Test or to find a Confidence Interval

7. 7 Tests for Two Dependent Means Goal: Compare the mean of the differences H 0 : μ d = 0 H 1 : μ d ≠ 0 Two tailedLeft tailedRight tailed H 0 : μ d = 0 H 1 : μ d < 0 H 0 : μ d = 0 H 1 : μ d > 0

8. 8 t =t = d – µ d sdsd n degrees of freedom: df = n – 1 Note:  d = 0 according to H 0 Finding the Test Statistic

9. 9 Test Statistic Note: Hypothesis Tests are done in same way as in Ch.8-5 Degrees of freedom df = n – 1

10. 10 Steps for Performing a Hypothesis Test on Two Independent Means Write what we know State H 0 and H 1 Draw a diagram Calculate the Sample Stats Find the Test Statistic Find the Critical Value(s) State the Initial Conclusion and Final Conclusion Note: Same process as in Chapter 8

11. 11 Example 1 Assume the differences in weight form a normal distribution. Use a 0.05 significance level to test the claim that for the population of students, the mean change in weight from September to April is 0 kg (i.e. on average, there is no change) Claim: μ d = 0 using α = 0.05

12. 12 H0 : µd = 0H1 : µd ≠ 0H0 : µd = 0H1 : µd ≠ 0 t = 0.186 t α/2 = 2.78 t-dist. df = 4 Test Statistic Critical Value Initial Conclusion: Since t is not in the critical region, accept H 0 Final Conclusion: We accept the claim that mean change in weight from September to April is 0 kg. -t α/2 = -2.78 Example 1 t α/2 = t 0.025 = 2.78 (Using StatCrunch, df = 4) d Data: -1 -1 4 -2 1 Sample Stats n = 5 d = 0.2 s d = 2.387 Use StatCrunch: Stat – Summary Stats – Columns Two-Tailed H 0 = Claim

13. 13 H0 : µd = 0H1 : µd ≠ 0H0 : µd = 0H1 : µd ≠ 0 Two-Tailed H 0 = Claim Initial Conclusion: Since P-value is greater than α (0.05), accept H 0 Final Conclusion: We accept the claim that mean change in weight from September to April is 0 kg. Example 1 d Data: -1 -1 4 -2 1 Sample Stats n = 5 d = 0.2 s d = 2.387 Use StatCrunch: Stat – Summary Stats – Columns Null: proportion= Alternative Sample mean: Sample std. dev.: Sample size: ● Hypothesis Test 0.2 2.387 5 0≠0≠ P-value = 0.8605 Stat → T statistics→ One sample → With summary

14. 14 Confidence Interval Estimate We can observe how the two proportions relate by looking at the Confidence Interval Estimate of μ 1 –μ 2 CI = ( d – E, d + E )

15. 15 Example 2 Find the 95% Confidence Interval Estimate of μ d from the data in Example 1 Sample Stats n = 5 d = 0.2 s d = 2.387 CI = (-2.8, 3.2) t α/2 = t 0.025 = 2.78 (Using StatCrunch, df = 4)

16. 16 Example 2 Find the 95% Confidence Interval Estimate of μ d from the data in Example 1 Sample Stats n = 5 d = 0.2 s d = 2.387 Level: Sample mean: Sample std. dev.: Sample size: ● Confidence Interval 0.2 2.387 5 0.95 Stat → T statistics→ One sample → With summary CI = (-2.8, 3.2)

17. 17

18. 18 Objective Compare of two population variances using two samples from each population. Hypothesis Tests and Confidence Intervals of two variances use the F-distribution Section 9.5 Comparing Variation in Two Samples

19. 19 (1) The two populations are independent (2) The two samples are random samples (3) The two populations are normally distributed (Very strict!) Requirements All requirements must be satisfied to make a Hypothesis Test or to find a Confidence Interval

20. 20 Important The first sample must have a larger sample standard deviation s 1 than the second sample. i.e. we must have s 1 ≥ s 2 If this is not so, i.e. if s 1 < s 2, then we will need to switch the indices 1 and 2

21. 21 Notation σ 1 First population standard deviation s 1 First sample standard deviation n 1 First sample size σ 2 Second population standard deviation s 2 Second sample standard deviation n 2 Second sample size Note: Use index 1 on sample/population with the larger sample standard deviation (s)

22. 22 Tests for Two Proportions H 0 : σ 1 = σ 2 H 1 : σ 1 ≠ σ 2 Two tailedRight tailed Note: We do not consider σ 1 < σ 2 (since we used indexes 1 and 2 such that s 1 is larger) Note: We only test the relation between σ 1 and σ 2 (not the actual numerical values) The goal is to compare the two population variances (or standard deviations) H 0 : σ 1 = σ 2 H 1 : σ 1 > σ 2

23. 23 The F-Distribution Similar to the χ 2 -dist. Not symmetric Non-negative values (F ≥ 0) Depends on two degrees of freedom df 1 = n 1 – 1 (Numerator df ) df 2 = n 2 – 1 (Denominator df )

24. 24 The F-Distribution On StatCrunch: Stat – Calculators – F df 1 = n 1 – 1 df 2 = n 2 – 1

25. 25 s1 s1 F = s2 s2 2 2 Test Statistic for Hypothesis Tests with Two Variances Where s 1 2 is the first (larger) of the two sample variances Because of this, we will always have F ≥ 1

26. 26 If the two populations have equal variances, then F = s 1 2 /s 2 2 will be close to 1 (Since s 1 2 and s 2 2 will be close in value) If the two populations have different variances, then F = s 1 2 /s 2 2 will be greater than 1 (Since s 1 2 will be larger than s 2 2 ) Use of the F Distribution

27. 27 Conclusions from the F-Distribution Values of F close to 1 are evidence in favor of the claim that the two variances are equal. Large values of F, are evidence against this claim (i.e. it suggest there is some difference between the two)

28. 28 Steps for Performing a Hypothesis Test on Two Independent Means Write what we know Index the variables such that s 1 ≥ s 2 (important!) State H 0 and H 1 Draw a diagram Find the Test Statistic Find the two degrees of freedom Find the Critical Value(s) State the Initial Conclusion and Final Conclusion

29. 29 Example 1 Below are sample weights (in g) of quarters made before 1964 and weights of quarters made after 1964. When designing coin vending machines, we must consider the standard deviations of pre-1964 quarters and post-1964 quarters. Use a 0.05 significance level to test the claim that the weights of pre-1964 quarters and the weights of post-1964 quarters are from populations with the same standard deviation. Claim: σ 1 = σ 2 using α = 0.05

30. 30 H 0 : σ 1 = σ 2 H 1 : σ 1 ≠ σ 2 Test Statistic Critical Value Initial Conclusion: Since F is in the critical region, reject H 0 Final Conclusion: We reject the claim that the weights of the pre-1964 and post-1964 quarters have the same standard deviation Example 1 Using StatCrunch: Stat – Calculators – F F α/2 = F 0.025 = 1.891 Two-Tailed H 0 = Claim n 1 = 40 n 2 = 40 α = 0.05 s 1 = 0.08700 s 2 = 0.06194 (Note: s 1 ≥s 2 ) 2 2 Degrees of Freedom df 1 = n 1 – 1 = 39 df 2 = n 2 – 1 = 39 F α/2 = 1.891 F = 1.973 F is in the critical region

31. 31 H 0 : σ 1 = σ 2 H 1 : σ 1 ≠ σ 2 Initial Conclusion: Since P-value is less than α (0.05), reject H 0 Final Conclusion: We reject the claim that the weights of the pre-1964 and post-1964 quarters have the same standard deviation Example 1 Two-Tailed H 0 = Claim n 1 = 40 n 2 = 40 α = 0.05 s 1 = 0.08700 s 2 = 0.06194 (Note: s 1 ≥s 2 ) Null: variance ratio= Alternative Sample 1: Variance: Size: Sample 2: Variance: Size: ● Hypothesis Test 0.007569 P-value = 0.0368 Stat → Variance → Two sample → With summary s 1 2 = 0.007569 s 2 2 = 0.003837 0.003837 40 1≠1≠