Objective Section 9.4 Inferences About Two Means (Matched Pairs) 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 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. 1 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. 1 First sample: weights of students in April Second sample: their weights in September These weights make 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. 3 4 d μ d n d s d Notation Individual difference between two matched paired values Population mean for the difference of the two values. Number of paired values in sample Mean value of the differences in sample Standard deviation of differences in sample Requirements (1) The sample data are dependent (i.e. they make matched pairs) () Either or both the following holds: The number of matched pairs is large (n>30) or The differences have a normal distribution All requirements must be satisfied to make a Hypothesis Test or to find a Confidence Interval 6
Tests for Two Dependent Means Goal: Compare the mean of the differences H 0 : μ d = 0 H 1 : μ d 0 Two tailed H 0 : μ d = 0 H 1 : μ d < 0 Left tailed H 0 : μ d = 0 H 1 : μ d > 0 Right tailed Finding the Test Statistic t = d µ d s d n Note: m d = 0 according to H 0 degrees of freedom: df = n 1 7 8 Test Statistic Degrees of freedom df = n 1 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 Find the Test Statistic Find the Critical Value(s) State the Initial Conclusion and Final Conclusion Note: Hypothesis Tests are done in same way as in Ch.8-9 Note: Same process as in Chapter 8 10 Assume the differences in weight form a normal distribution. Use a 0.0 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) H 0 : µ d = 0 H 1 : µ d 0 Test Statistic n = d = 0. s d =.387 d Data: -1-1 4-1 Use StatCrunch: Stat Summary Stats Columns -t α/ = -.78 Critical Value t α/ = t 0.0 =.78 (Using StatCrunch, df = 4) t = 0.186 t-dist. df = 4 t α/ =.78 Claim: μ d = 0 using α = 0.0 11 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. 1
d Data: -1-1 4-1 H 0 : µ d = 0 H 1 : µ d 0 n = d = 0. s d =.387 Use StatCrunch: Stat Summary Stats Columns Stat T statistics One sample With summary Sample mean: Sample std. dev.: Sample size: 0..387 Hypothesis Test Null: proportion= Alternative P-value = 0.860 0 Confidence Interval Estimate We can observe how the two proportions relate by looking at the Confidence Interval Estimate of μ 1 μ CI = ( d E, d + E ) Initial Conclusion: Since P-value is greater than α (0.0), accept H 0 Final Conclusion: We accept the claim that mean change in weight from September to April is 0 kg. 13 14 Example Find the 9% Confidence Interval Estimate of μ d from the data in n = d = 0. s d =.387 t α/ = t 0.0 =.78 (Using StatCrunch, df = 4) 3 Example Find the 9% Confidence Interval Estimate of μ d from the data in n = d = 0. s d =.387 Stat T statistics One sample With summary Sample mean: Sample std. dev.: Sample size: 0..387 Confidence Interval Level: 0.9 CI = (-.8, 3.) CI = (-.8, 3.) 1 16 Section 9. Comparing Variation in Two Samples Objective Compare of two population variances using two samples from each population. Hypothesis Tests and Confidence Intervals of two variances use the F-distribution 17 18
Requirements (1) The two populations are independent () The two samples are random samples (3) The two populations are normally distributed (Very strict!) 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 All requirements must be satisfied to make a Hypothesis Test or to find a Confidence Interval If this is not so, i.e. if s 1 < s, then we will need to switch the indices 1 and 19 0 Notation Tests for Two Proportions σ 1 s 1 First population standard deviation First sample standard deviation The goal is to compare the two population variances (or standard deviations) n 1 σ First sample size Second population standard deviation 4 H 0 : σ 1 = σ H 1 : σ 1 σ H 0 : σ 1 = σ H 1 : σ 1 > σ s Second sample standard deviation Two tailed Right tailed n Second sample size Note: Use index 1 on sample/population with the larger sample standard deviation (s) 1 Note: We do not consider σ 1 < σ (since we used indexes 1 and such that s 1 is larger) Note: We only test the relation between σ 1 and σ (not the actual numerical values) The F-Distribution The F-Distribution Similar to the χ -dist. Not symmetric df 1 = n 1 1 df = n 1 Non-negative values (F 0) Depends on two degrees of freedom df 1 = n 1 1 (Numerator df ) df = n 1 (Denominator df ) On StatCrunch: Stat Calculators F 3 4
Test Statistic for Hypothesis Tests with Two Variances s 1 F = s Use of the F Distribution If the two populations have equal variances, then F = s 1 /s will be close to 1 (Since s 1 and s will be close in value) Where s 1 is the first (larger) of the two sample variances Because of this, we will always have F 1 If the two populations have different variances, then F = s 1 /s will be greater than 1 (Since s 1 will be larger than s ) 6 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) Steps for Performing a Hypothesis Test on Two Independent Means Write what we know Index the variables such that s 1 s (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 7 8 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 0.0 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. H 0 : σ 1 = σ H 1 : σ 1 σ Test Statistic Critical Value Using StatCrunch: Stat Calculators F F α/ = F 0.0 = 1.891 Degrees of Freedom df 1 = n 1 1 = 39 df = n 1 = 39 n 1 = 40 n = 40 α = 0.0 s 1 = 0.08700 s = 0.06194 (Note: s 1 s ) F α/ = 1.891 F = 1.973 F is in the critical region Claim: σ 1 = σ using α = 0.0 9 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 30
H 0 : σ 1 = σ H 1 : σ 1 σ n 1 = 40 n = 40 α = 0.0 s 1 = 0.08700 s = 0.06194 (Note: s 1 s ) s 1 = 0.00769 s = 0.003837 Stat Variance Two sample With summary Sample 1: Variance: 0.00769 Hypothesis Test Size: 40 Null: variance ratio= Sample : Variance: 0.003837 Alternative Size: 40 1 P-value = 0.0368 Initial Conclusion: Since P-value is less than α (0.0), 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 31 6