Question

Test the claim that the proportion of people who own cats is smaller than 50% at the 0.01 significance level. The null and alternative hypothesis would be: Họ: A = 0.5 Ho:p> 0.5 Hp p = 0.5 HP < 0.5 Hp: < 0.5 Hp:u > 0.5 H:+0.5 H :p <0.5 H :P +0.5 H :p > 0.5 H :p > 0.5 H :< 0.5 O O O The test is: right-tailed two-tailed left-tailed Based on a sample of 500 people, 48% owned cats The p-value is: (to 2 decimals) Based on this we: O Reject the null hypothesis Fail to reject the null hypothesis

Answer 1

The null and alternate hypothesis are:

H0: $p\geq 0.5$

H1:

Hence 2nd option.

This is a left tailed test.

The test statistic is given by:

$Z=\frac{\widehat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}=\frac{0.48-0.5}{\sqrt{\frac{0.5(1-0.5)}{500}}}=-0.89$

Since this is a left-tailed test, so the p-value is given by:

$\\P(Z<-0.89)=P(Z>0.89)=1-P(Z<0.89)=1-0.81327\\ \\=0.18673\approx 0.19$

Since p-value is greater than 0.01, so we do not have sufficient evidence to reject the null hypothesis H0.

Thus, we fail to reject the null hypothesis.