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# 06 1645 t005899 so reject the null finally state your

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Unformatted text preview: red to have a weekly labor supply (hours worked) of 30 hours. The standard deviation is calculated to be s = 10. You believe that in the population mean hours worked per week is less than 32 hours/week. Perform a test to support your idea. Step 1: The null and alternative hypotheses are: H0 :µ ≥ 32 H1 :µ &lt; 32 Utku Suleymanoglu (UMich) Hypothesis Testing 29 / 37 σ not Known Step 2: Calculate the relevant test statistic: t= x − µ0 ¯ 30 − 32 −2 √= = = −6.06 10/30 0.33 s/ n Step 3: Find the critical value: Let’s pick α = 0.05. Then tα,n−1 = t0.05,899 ≈ 1.645 (Table 2 in Appendix B) (n is too large, use inﬁnity as df). Step 4: Use the decision rule: Reject H0 if t &lt; −tα,n−1 . t = −6.06 &lt; −1.645 ≈ −t0.05,899 , so reject the null. Finally, state your conclusion: We have enough evidence to reject the claim that µ ≥ 32. So our claim that it is less than 32 is supported by the data. Utku Suleymanoglu (UMich) Hypothesis Testing 30 / 37 σ not Known Two-Tailed Tests: σ Unknown We will again use t-distribution instead of standard normal. First state the null and alternative hypotheses. H 0 : µ = µ0 H 1 : µ = µ0 We reject the null the hypothesis if test statistic (same as one-tailed tests): t= x − µ0 ¯ √ s/ n is such that |t | &gt; tα/2,n−1 where tα/2 is the critical value with probability α/2 in the upper tail. So if test statistic falls outside the interval (−tα/2 , tα/2 ), the null is rejected. Utku Suleymanoglu (UMich) Hypothesis Testing 31 / 37 σ not Known Example Remember the previous data set that consists of 900 people aged 65. They are measured to have a weekly labor supply (hours worked) of 30 hours. The standard deviation is calculated to be s = 10. Now, somebody claims that mean hours worked is exactly 29.5. Test this claim. Step 1: The null and alternative hypotheses are: H0 :µ = 29.5 H1 :µ = 29.5 Utku Suleymanoglu (UMich) Hypothesis Testing 32 / 37 σ not Known Step 2: Calculate the relevant test statistic: t= x − µ0 ¯ 0.5 30 − 29.5 √= = = 1.515 10/30 0.33 s/ n Step 3: Find the critical value: Let’s pick α = 0.05. Then tα/2,n−1 = t0.025,899 ≈ 1.96 (Table 2 in Appendix B) (n is too large, use inﬁnity as df). Step 4: Use the decision rule: Reject H0 if |t | &gt; tα/2,n−1 . t = 1.515 is outside the rejection region (−1.96, 1.96). We failed to reject the null. As a ﬁnal step, state your conclusion: We do NOT have enough evidence to reject the claim that µ = 29.5. Note that this does not mean that we know that µ = 29.5. We would not have rejected the null with 29.6 or 30 either! Utku Suleymanoglu (...
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## This note was uploaded on 03/17/2014 for the course ECON 404 taught by Professor Staff during the Spring '08 term at University of Michigan.

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