Asked by Amy
Use a 1.15 level of significance to test a claim that a true proportion is less than 47% when n=250 and x-112
Answers
Answered by
Kuai
proportion = 112/250 = .448
z = (.448 - .47) / sqrt(.47*.53/250) = -0.70
p-value = .2420
Since the p-value is greater than .05, do not reject the null hypothesis. The claim is not supported.
z = (.448 - .47) / sqrt(.47*.53/250) = -0.70
p-value = .2420
Since the p-value is greater than .05, do not reject the null hypothesis. The claim is not supported.
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