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Lesson 10 - 2. Testing Claims about a Population Mean Assuming the Population Standard Deviation is Known. Objectives. Understand the logic of hypothesis testing Test a claim about a population mean with σ known using the classical approach
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Lesson 10 - 2 Testing Claims about a Population Mean Assuming the Population Standard Deviation is Known
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
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
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 H0 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
-zα/2 zα zα/2 -zα Classical Approach Critical Regions x – μ0 Test Statistic: z0 = ------------- σ/√n
-|z0| |z0| z0 z0 x – μ0 Test Statistic: z0 = ------------- σ/√n P-Value Approach P-Value is thearea highlighted
P-Value Examples For each α and observed significance level (p-value) pair, indicate whether the null hypothesis would be rejected. • α = . 05, p = .10 • α = .10, p = .05 • α = .01 , p = .001 • α = .025 , p = .05 e) α = .10, p = .45 α < P fail to reject Ho P < α reject Ho P < α reject Ho α < P fail to reject Ho α < P fail to reject Ho
Confidence Interval Approach Confidence Interval: x – zα/2 · σ/√n x + zα/2 · σ/√n Lower Bound Upper Bound μ0
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.
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. not equal two-tailed X-bar – μ 65.014 – 50.64 14.374 Z0 = --------------- = ---------------------- = ------------- = 2.69 σ / √n 18.49/√12 5.3376 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 z0 lies in the rejection region, we would reject H0. Zc = 1.96 Zc (α/2 = 0.025) = 1.96
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. not equal two-tailed X-bar – μ 65.014 – 50.64 14.374 Z0 = --------------- = ---------------------- = ------------- = 2.69 σ / √n 18.49/√12 5.3376 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 H0 because the p-value is less than α. -Z0 = -2.69 P( z < Z0 = -2.69) = 0.0036
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 μ0, 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:z0 = 2.693 and p-value = 0.0071
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. Confidence Interval = Point Estimate ± Margin of Error = μ ± Zα/2 σ / √n = 50.64 ± 1.96 (18.49) / √12 = 50.64 ± 10.4617 Zc (α/2) = 1.96 65.014 μ 40.18 61.10 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 H0.
Using Your Calculator: Z-Test From previous problem:u0 = 50.64 and interval (54.552, 75.476) Therefore Reject • 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 μ0 is in the interval, then FTR • If μ0 is outside the interval, then REJ
Summary and Homework • Summary • A hypothesis test of means compares whether the true mean is either • Equal to, or not equal to, μ0 • Equal to, or less than, μ0 • Equal to, or more than, μ0 • 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