According to a study by the American Pet Food Dealers Association, 63% of U.S. h

Question

According to a study by the American Pet Food Dealers Association, 63% of U.S. households own pets. A report is being prepared for an editorial in the San Francisco Chronicle. As a part of the editorial, a random sample of 300 households showed that 210 own pets. Do these data disagree with the Pet Food Dealers Association data? Use the 0.05 level of significance. a. State the null and alternative hypothesis. b. What is the level of significance? c. Which distribution (normal or t) should be used for this problem? Prove that this distribution can be used. d. Create a decision rule for this problem. State the critical value. e. Calculate the test statistic. f State the conclusion in detail. g. Calculate the p-value. Restate the conclusion.

Explanation / Answer

We are given that population proportion ,P= 0.63

and sample proportion,p = 210/300 = 0.7

Hypothesis is

Null hypothesis: P = 0.63
Alternative hypothesis: P 0.63

now the test statisitic is calculated as

= sqrt[ P * ( 1 - P ) / n ] = sqrt(0.63*(1-0.63)/300) = 0.0278

and z = (p - P) / = (0.70-0.63)/0.0278 = 2.51

now we check the z critical value for 0.05 level of significane , which is 1.96

as our z stat > z critical hence we reject the null hypothesis in favor of alternate hypothesis and conclude that P 0.70

using the z table again we see that for 2.51 the p value is 0.0120 , again the p value is less than 0.05 , hence we reject the null hypothesis in favor of alternate hypothesis and conclude that P 0.70

Hope this helps !!

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