SOLUTION: A department store manager claims that at least 45% of persons who visited this store make a purchase. In a sample of 400 persons who visited the store 150 made a purchase. Test

Question 247403: A department store manager claims that at least 45% of persons who visited this store make a purchase. In a sample of 400 persons who visited the store 150 made a purchase. Test the manager�s claim at the 3% significance level.
a.State the null and alternate hypothesis.
b. Determine the rejection region for the decision rule.
c. which equation listed below would you use?
d. Using the statistical calculator or manual calculation what is the observed statistic and what is your decision regarding the null hypothesis. Provide a one sentence answer.

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
A department store manager claims that at least 45% of persons who visited this store make a purchase. In a sample of 400 persons who visited the store 150 made a purchase. Test the manager�s claim at the 3% significance level.
a.State the null and alternate hypothesis.
Ho: p >= 0.45 (Claim)
H1: p < 0.45
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sample proportion = 150/400 = 0.375
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test statistic: z(0.375) = (0.375-0.45)/sqrt[0.45*0.55/400] = -3.0151...
critical value for left-tail test with alpha = 3%: z = -1.8808
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Since the test stat value is in the reject interval, reject Ho.
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b. Determine the rejection region for the decision rule.
critical value for left-tail test with alpha = 3%: z = -1.8808
The reject interval is all z-values less than -1.8808
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c. which equation listed below would you use?
The equations are not listed.
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d. Using the statistical calculator or manual calculation what is the observed statistic and what is your decision regarding the null hypothesis. Provide a one sentence answer.
Answered above.
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Cheers,
Stan H.

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