Math119SolsFinalS11

Math119SolsFinalS11 - SOLUTIONS TO THE FINAL MATH 119...

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SOLUTIONS TO THE FINAL MATH 119 – SPRING 2011 – KUNIYUKI 126 POINTS TOTAL, BUT 120 POINTS = 100% When rounding , round off to at least four decimal places or four significant digits, whichever is more detailed . Round off final z values to two decimal places. Round off final t and ± 2 values to three decimal places. Assume that finite population correction factors do not apply. Do not use continuity corrections. FOR PROBLEM 1, USE THE P -VALUE METHOD OF HYPOTHESIS TESTING. Remember to: • State the null and alternative hypotheses using notation (as in class), and identify which is the claim. • Compute the value of the appropriate test statistic. • Give the corresponding P -value. • State whether or not the null hypothesis is rejected; this is your “decision.” • Write your final conclusion relative to the claim using the kind of wording we used in class. 1) (19 points). The average income of adults living in Capital City last year was $38,000, and the standard deviation of such incomes was $6000. We take a random sample of 60 adults living in Capital City who graduated from Capital City College (CCC), and their average income last year was $40,000. Use a significance level of 0.10 to test the claim that last year’s average income of CCC graduates living in Capital City was equal to $38,000. Assume that the standard deviation of the incomes of CCC graduates living in Capital City last year was $6000, even though this may be a risky assumption. Use the z table and the P -value method of hypothesis testing. Note : The population standard deviation is actually probably less than $6000, since CCC graduates have at least their CCC degrees in common. Because we are testing for a population mean, and the sample size n = 60 > 30 , the CLT applies. Furthermore, is presumed known, so we use the z test statistic. Let μ = last year’s average income of CCC graduates living in Capital City (in dollars).
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State H 0 , H A , ± : H 0 : μ = $38,000 Claim () H A : ± $38,000 ² Two-tailed test () ³ = 0.10 Test statistic: z = x ± ² n = 40,000 ± 38,000 6000 60 ³ 2000 774.5967 ³ 2.58 or z = x ± x x where x = = 38,000 and x = n = 6000 60 ³ 774.5967 ³ 40,000 ± 38,000 774.5967 ³ 2000 774.5967 ³ 2.58 Find the corresponding P -value (remember, we have a two-tailed test here). P -value ± 2 .0049 () = .0098 Decision: P -value < , because .0098 < .10 , so the P -value is low. We reject
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This note was uploaded on 09/08/2011 for the course MATH 119 taught by Professor Staff during the Spring '11 term at Mesa CC.

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Math119SolsFinalS11 - SOLUTIONS TO THE FINAL MATH 119...

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