a) The following null and alternative hypotheses need to be tested:
"H_1: p>0.5"
b) The z-statistic is computed as follows:
Let "n=200, x=115"
Then
"z=\\dfrac{0.575-0.5}{\\sqrt{\\dfrac{0.5(1-0.5)}{200}}}\\approx2.1213"
Using the P-value approach: The p-value is "p=P(Z>2.1213)=0.016948," and since "p=0.016948<0.05=\\alpha," it is concluded that the null hypothesis is rejected.
c) There is enough evidence to claim that the population proportion "p" is greater than "p_0=0.5," at the "\\alpha=0.05" significance level.
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