Question

A recent article in a reputable journal states that that 38% of the consumer scam complaints by citizens of a certain country are about identity theft. Suppose a random sample of 94 complaints is obtained. Of these complaints, 39 were regarding identity theft. Based on these sample data, what conclusion should be reached about the statement made in the journal? Test using α-0.10. Determine the null and alternative hypotheses. Select the correct choice below and fill in the answer boxes within your choice. Type integers or decimals.) Check the requirement for a hypothesis test for a proportion In this situation, пр- I is Type integers or decimals.) What is the p-value? p-value Round to three decimal places as needed.) What is the proper conclusion? ▼ | 5 and n( 1-p)-D is ▼5. Thus, the requirement | ▼| satisfied. Ho.There is ▼ evidence that the 38% rate quoted in the journal article isHow do I find the p-value in excel?

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Answer #1

(a) Ho: π = 0.38 and Ha: π ≠ 0.38

(b) np = 39 is > 5 and n(1 - p) = 55 > 5. Thus the requirement is satisfied.

(c)

Data:    

n = 94   

p = 0.38   

p' = 0.414893617   

Hypotheses:    

Ho: p = 0.38   

Ha: p ≠ 0.38   

Decision Rule:    

α = 0.1   

Lower Critical z- score = -1.6449   

Upper Critical z- score = 1.6449   

Reject Ho if |z| > 1.6449   

Test Statistic:    

SE = √{p (1 - p)/n} = √(0.38 * (1 - 0.38)/94) = 0.0501

z = (p'- p)/SE = (0.414893617021277 - 0.38)/0.0500637890967454 = 0.6970

p- value = 0.486

Decision (in terms of the hypotheses):    

Since 0.6970 < 1.6449 we fail to reject Ho

Conclusion (in terms of the problem):    

There is no sufficient evidence that 38% rate quoted in the journal is not valid.   

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