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(1 proportion hypothesis testing) I read in a newspaper that 37% of Americans dr

ID: 3319390 • Letter: #

Question

(1 proportion hypothesis testing) I read in a newspaper that 37% of Americans drink tea at least once a week. I want to test this claim Let's set the null hypothesis to be p'-0.37 and the alternate hypothesis to be p'*0.37. (a) For this hypothesis test, what would a type l error mean? (b) I ask 100 people and find that 43 of them drink tea at least once a week. What kind of hypothesis test should we perform? Perform the hypothesis test at the 95% confidence level, and state the p-value (c) Should we accept or reject the null hypothesis? State the conclusion of this study with respect to the claim, in clear simple language.

Explanation / Answer

a) Reject Null hypothesis, when P = 0.37 is true

b) Here Z test for single porportion is suitable

95% confidence interval =

= (p ± Z critical * pq/n)

= (0.43 ± 1.96 * 0.43*0.57/100)

= ( 0.333, 0.527)

c) Accept null hypothesis, since the above interval contains 0.37