Survey

Survey

Document related concepts

no text concepts found

Transcript

Section 9.4 Inferences About Two Means (Matched Pairs) 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 1 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. 2 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. 3 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. 4 Notation d Individual difference between two matched paired values μd Population mean for the difference of the two values. n Number of paired values in sample d Mean value of the differences in sample sd Standard deviation of differences in sample 5 Requirements (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 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 H0 : μd = 0 H0 : μd = 0 H0 : μd = 0 H1 : μd ≠ 0 H1 : μd < 0 H1 : μd > 0 Two tailed Left tailed Right tailed 7 Finding the Test Statistic t= d – µd sd n Note: md = 0 according to H0 degrees of freedom: df = n – 1 8 Test Statistic Degrees of freedom df = n – 1 Note: Hypothesis Tests are done in same way as in Ch.8-5 9 Steps for Performing a Hypothesis Test on Two Independent Means • Write what we know • State H0 and H1 • 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 10 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: μd = 0 using α = 0.05 11 Example 1 d Data: -1 -1 4 -2 1 H0 : µd = 0 H1 : µd ≠ 0 Two-Tailed H0 = Claim n=5 d = 0.2 t = 0.186 -tα/2 = -2.78 Sample Stats t-dist. df = 4 tα/2 = 2.78 sd = 2.387 Use StatCrunch: Stat – Summary Stats – Columns Test Statistic Critical Value tα/2 = t0.025 = 2.78 (Using StatCrunch, df = 4) Initial Conclusion: Since t is not in the critical region, accept H0 Final Conclusion: We accept the claim that mean change in weight from September to April is 0 kg. 12 Example 1 d Data: -1 -1 4 -2 1 Sample Stats H0 : µd = 0 H1 : µd ≠ 0 n=5 Two-Tailed H0 = Claim d = 0.2 sd = 2.387 Use StatCrunch: Stat – Summary Stats – Columns Stat → T statistics→ One sample → With summary Sample mean: 0.2 Sample std. dev.: 2.387 Sample size: 5 ● Hypothesis Test Null: proportion= 0 Alternative ≠ P-value = 0.8605 Initial Conclusion: Since P-value is greater than α (0.05), accept H0 Final Conclusion: We accept the claim that mean change in weight from September to April is 0 kg. 13 Confidence Interval Estimate We can observe how the two proportions relate by looking at the Confidence Interval Estimate of μ1–μ2 CI = ( d – E, d + E ) 14 Example 2 Sample Stats n=5 d = 0.2 tα/2 = t0.025 = 2.78 Find the 95% Confidence Interval Estimate of μd from the data in Example 1 sd = 2.387 (Using StatCrunch, df = 4) CI = (-2.8, 3.2) 15 Example 2 Sample Stats n=5 d = 0.2 Find the 95% Confidence Interval Estimate of μd from the data in Example 1 sd = 2.387 Stat → T statistics→ One sample → With summary Sample mean: 0.2 Sample std. dev.: 2.387 Sample size: 5 ● Confidence Interval Level: 0.95 CI = (-2.8, 3.2) 16 17 Section 9.5 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 18 Requirements (1) The two populations are independent (2) The two samples are random samples (3) The two populations are normally distributed (Very strict!) All requirements must be satisfied to make a Hypothesis Test or to find a Confidence Interval 19 Important The first sample must have a larger sample standard deviation s1 than the second sample. i.e. we must have s1 ≥ s2 If this is not so, i.e. if s1 < s2 , then we will need to switch the indices 1 and 2 20 Notation σ1 First population standard deviation s1 First sample standard deviation n1 First sample size σ2 Second population standard deviation s2 Second sample standard deviation n2 Second sample size Note: Use index 1 on sample/population with the larger sample standard deviation (s) 21 Tests for Two Proportions The goal is to compare the two population variances (or standard deviations) H0 : σ1 = σ2 H0 : σ1 = σ2 H1 : σ1 ≠ σ2 H1 : σ1 > σ2 Two tailed Right tailed Note: We do not consider σ1 < σ2 (since we used indexes 1 and 2 such that s1 is larger) Note: We only test the relation between σ1 and σ2 (not the actual numerical values) 22 The F-Distribution Similar to the χ2-dist. • Not symmetric • Non-negative values (F ≥ 0) • Depends on two degrees of freedom df1 = n1 – 1 (Numerator df ) df2 = n2 – 1 (Denominator df ) 23 The F-Distribution df1 = n1 – 1 df2 = n2 – 1 On StatCrunch: Stat – Calculators – F 24 Test Statistic for Hypothesis Tests with Two Variances F= s s 2 1 2 2 Where s12 is the first (larger) of the two sample variances Because of this, we will always have F ≥ 1 25 Use of the F Distribution If the two populations have equal variances, then F = s12/s22 will be close to 1 (Since s12 and s22 will be close in value) If the two populations have different variances, then F = s12/s22 will be greater than 1 (Since s12 will be larger than s22) 26 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) 27 Steps for Performing a Hypothesis Test on Two Independent Means • Write what we know • Index the variables such that s1 ≥ s2 (important!) • State H0 and H1 • 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 28 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: σ1 = σ2 using α = 0.05 29 Example 1 H0 : σ1 = σ2 H1 : σ1 ≠ σ2 n1 = 40 n2 = 40 s1 = 0.08700 s2 = 0.06194 α = 0.05 (Note: s1≥s2) Two-Tailed H0 = Claim F = 1.973 Test Statistic 2 Fα/2 = 1.891 2 Degrees of Freedom df1 = n1 – 1 = 39 df2 = n2 – 1 = 39 Critical Value F is in the critical region Using StatCrunch: Stat – Calculators – F Fα/2 = F0.025 = 1.891 Initial Conclusion: Since F is in the critical region, reject H0 Final Conclusion: We reject the claim that the weights of the pre-1964 and post-1964 quarters have the same standard deviation 30 Example 1 H0 : σ1 = σ2 H1 : σ1 ≠ σ2 n1 = 40 n2 = 40 s1 = 0.08700 s2 = 0.06194 s12 = 0.007569 Two-Tailed H0 = Claim α = 0.05 (Note: s1≥s2) s22 = 0.003837 Stat → Variance → Two sample → With summary Sample 1: Variance: 0.007569 40 Size: Sample 2: Variance: 0.003837 Size: ● Hypothesis Test Null: variance ratio= 1 Alternative ≠ 40 P-value = 0.0368 Initial Conclusion: Since P-value is less than α (0.05), reject H0 Final Conclusion: We reject the claim that the weights of the pre-1964 and post-1964 quarters have the same standard deviation 31

Related documents