1. Lesson 10 - 2 Testing Claims about a Population Mean Assuming the Population Standard Deviation is Known

2. Objectives Understand the logic of hypothesis testing Test a claim about a population mean with σ known using the classical approach Test a claim about a population mean with σ known using P-values Test a claim about a population mean with σ known using confidence intervals Understand the difference between statistical significance and practical significance

3. Vocabulary Statistically Significant – when observed results are unlikely under the assumption that the null hypothesis is true. When results are found to be statistically significant, we reject the null hypothesis Practical Significance – refers to things that are statistically significant, but the actual difference is not large enough to cause concern or be considered important

4. Hypothesis Testing Approaches Classical –Logic: If the sample mean is too many standard deviations from the mean stated in the null hypothesis, then we reject the null hypothesis (accept the alternative) P-Value –Logic: Assuming H 0 is true, if the probability of getting a sample mean as extreme or more extreme than the one obtained is small, then we reject the null hypothesis (accept the alternative). Confidence Intervals –Logic: If the sample mean lies in the confidence interval about the status quo, then we fail to reject the null hypothesis

5. zαzα -z α/2 z α/2 -z α Critical Regions x – μ 0 Test Statistic: z 0 = ------------- σ/√n Reject null hypothesis, if Left-TailedTwo-TailedRight-Tailed z 0 < - z α z 0 < - z α/2 or z 0 > z α/2 z 0 > z α Classical Approach

6. z0z0 -|z 0 | |z 0 | z0z0 P-Value is the area highlighted x – μ 0 Test Statistic: z 0 = ------------- σ/√n Reject null hypothesis, if P-Value < α P-Value Approach

7. P-Value Examples For each α and observed significance level (p-value) pair, indicate whether the null hypothesis would be rejected. a)α =. 05, p =.10 b)α =.10, p =.05 c)α =.01, p =.001 d)α =.025, p =.05 e) α =.10, p =.45 α < P  fail to reject H o P < α  reject H o

8. Reject null hypothesis, if μ 0 is not in the confidence interval Confidence Interval: x – z α/2 · σ/√n x + z α/2 · σ/√n Confidence Interval Approach Lower Bound Upper Bound μ0μ0

9. Example 1 A simple random sample of 12 cell phone bills finds x- bar = $65.014. The mean in 2004 was $50.64. Assume σ = $18.49. Test if the average bill is different today at the α = 0.05 level. Use each approach.

10. Example 1: Classical Approach A simple random sample of 12 cell phone bills finds x-bar = $65.014. The mean in 2004 was $50.64. Assume σ = $18.49. Test if the average bill is different today at the α = 0.05 level. Use the classical approach. X-bar – μ 65.014 – 50.64 14.374 Z 0 = --------------- = ---------------------- = ------------- = 2.69 σ / √n 18.49/√12 5.3376 not equal  two-tailed Z c = 1.96 Using alpha, α = 0.05 the shaded region are the rejection regions. The sample mean would be too many standard deviations away from the population mean. Since z 0 lies in the rejection region, we would reject H 0. Z c (α/2 = 0.025) = 1.96

11. Example 1: P-Value A simple random sample of 12 cell phone bills finds x-bar = $65.014. The mean in 2004 was $50.64. Assume σ = $18.49. Test if the average bill is different today at the α = 0.05 level. Use the P-value approach. X-bar – μ 65.014 – 50.64 14.374 Z 0 = --------------- = ---------------------- = ------------- = 2.69 σ / √n 18.49/√12 5.3376 not equal  two-tailed -Z 0 = -2.69 The shaded region is the probability of obtaining a sample mean that is greater than $65.014; which is equal to 2(0.0036) = 0.0072. Using alpha, α = 0.05, we would reject H 0 because the p-value is less than α. P( z < Z 0 = -2.69) = 0.0036

12. Using Your Calculator: Z-Test For classical or p-value approaches Press STAT –Tab over to TESTS –Select Z-Test and ENTER Highlight Stats Entry μ 0, σ, x-bar, and n from summary stats Highlight test type (two-sided, left, or right) Highlight Calculate and ENTER Read z-critical and/or p-value off screen From previous problem: z 0 = 2.693 and p-value = 0.0071

13. Example 1: Confidence Interval A simple random sample of 12 cell phone bills finds x-bar = $65.014. The mean in 2004 was $50.64. Assume σ = $18.49. Test if the average bill is different today at the α = 0.05 level. Use confidence intervals. The shaded region is the region outside the 1- α, or 95% confidence interval. Since the sample mean lies outside the confidence interval, then we would reject H 0. Confidence Interval = Point Estimate ± Margin of Error = μ ± Z α/2 σ / √n = 50.64 ± 1.96 (18.49) / √12 = 50.64 ± 10.4617 Z c (α/2) = 1.96 μ 65.014 40.1861.10

14. Using Your Calculator: Z-Test Press STAT –Tab over to TESTS –Select Z-Interval and ENTER Highlight Stats Entry σ, x-bar, and n from summary stats Entry your confidence level (1- α) Highlight Calculate and ENTER Read confidence interval off of screen –If μ 0 is in the interval, then FTR –If μ 0 is outside the interval, then REJ From previous problem: u 0 = 50.64 and interval (54.552, 75.476) Therefore Reject

15. Summary and Homework Summary –A hypothesis test of means compares whether the true mean is either Equal to, or not equal to, μ 0 Equal to, or less than, μ 0 Equal to, or more than, μ 0 –There are three equivalent methods of performing the hypothesis test The classical approach The P-value approach The confidence interval approach Homework –pg 526 – 530: 1, 3, 4, 10, 12, 17, 28, 29, 30