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Reporting Test Results asp-Values: How to Decide Whether to Reject H1.Choose the maximum value of that you are willing to tolerate.2.If the observed significance level (p-value) of the test is less than the chosen value of reject the null hypothesis. Otherwise, do not reject the null hypothesis.0,
Two-Tailed zTest p-Value Example Does an average box of cereal contain 368grams of cereal? random sample of 25boxes showed x= 372.5. The company has specified to be 15grams. Find the p-value. How does it compare to = .05?A
Two-Tailed zTest p-Value Solutionz01.50zvalue of sample statistic (observed)zxn372.536815251.50
Two-Tailed ZTest p-Value Solution1/2 p-Value1/2 p-Valuezvalue of sample statistic (observed)p-Value is P(z–1.50 or z1.50)z01.50–1.50From ztable: lookup 1.50.4332.5000– .4332.0668
Two-Tailed zTest p-Value Solution1/2 p-Value.06681/2 p-Value.0668p-Value is P(z–1.50 or z1.50) = .1336z01.50–1.50
Two-Tailed zTest p-Value Solution01.50–1.50zReject H0Reject H01/2 p-Value = .06681/2 p-Value = .06681/2 = .0251/2 = .025p-Value = .1336 = .05 Do not reject H0.Test statistic is in ‘Do not reject’ region
One-Tailed zTest p-Value Example Does an average box of cereal contain more than368grams of cereal? A random sample of 25boxes showed x= 372.5The company has specified to be 15grams. Find the value. How does it compare to = .05?. p-
One-Tailed zTest p-Value Solutionz01.50zvalue of sample statisticzxn372.536815251.50
One-Tailed zTest p-Value SolutionUse alternative hypothesis to find directionp-Value is P(z1.50) zvalue of sample statisticp-Valuez01.50From ztable: lookup 1.50.4332.5000– .4332.0668
One-Tailed zTest p-Value Solutionp-Value.0668zvalue of sample statisticFrom ztable: lookup 1.50Use alternative hypothesis to find direction.5000– .4332.0668p-Value is P(z1.50) = .0668z01.50.4332
= .05One-Tailed zTest p-Value Solution01.50zReject H0p-Value = .0668(p-Value = .0668) (= .05). Do not reject H0.Test statistic is in ‘Do not reject’ region
p-Value Thinking ChallengeYou’re an analyst for Ford. You want to find out if the average miles per gallon of Escorts is less than 32 mpg. Similar models have a standard deviation of 3.8mpg. You take a sample of 60Escorts & compute a sample mean of 30.7mpg. What is the p-value? How does it compare to = .01?
Use alternative hypothesis to find directionp-Value Solution*z0–2.65zvalue of sample statisticFrom ztable: lookup 2.65.4960p-Value.004.5000– .4960.0040p-Value is P(z-2.65) = .004.p-Value < (= .01). Reject H0.
Converting a Two-Tailedp-Value from a Printout to a One-Tailed p-Valueif Hais of the form > and zis positiveor Hais of the form < and zis negativeif Hais of the form > and zis negativeHais of the form < and zis positivepReported p-value2p1Reported p-value2
7.4Test of Hypotheses about a Population Mean:Normal (z) Statistic
Large-Sample Test of Hypothesis about µOne-Tailed TestTwo-Tailed TestH0: µ= µ0H0: µ= µ0Ha: µ< µ0Ha: µ≠ µ0(or Ha: µ> µ0)Test Statistic:Test Statistic:zxµ0xxµ0snzxµ0xxµ0sn
Large-Sample Test of Hypothesis about µOne-Tailed TestRejection region:z< –z(orz> zwhenHa: µ> µ0)where zis chosen so thatP(z> z) =