Question

Test the claim that the proportion of people who own cats is larger than 60% at the 0.005 significance level. The null and alternative hypothesis would be: Ho:p = 0.6 Ho:p > 0.6 Ho:4 > 0.6 Ho:u < 0.6 Ha:p + 0.6 Ha:p < 0.6 Ha:4 < 0.6 Ha:> 0.6 Ho:u = 0.6 Ho:p < 0.6 Haiu + 0.6 Ha:p > 0.6 The test is: left-tailed right-tailed two-tailed Based on a sample of 500 people, 65% owned cats The p-value is: (to 2 decimals) Based on this we: Fail to reject the null hypothesis Reject the null hypothesis

Answer 1

Solution :

This is the right tailed test .

The null and alternative hypothesis is

H₀ : p $\leq$ 0.60

H_a : p >0.60

$\hat p$ = 0.65

Test statistic = z

= $\hat p$ - P₀ / [$\sqrt$P_{0 *} (1 - P₀ ) / n]

= 0.65 - 0.60/ [$\sqrt$(0.60 * 0.40) / 500]

= 2.28

P(z > 2.28) = 1 - P(z < 2.28) = 0.0113

P-value = 0.0113 = 0.01

$\alpha$ = 0.005

P-value > $\alpha$

Fail to reject the null hypothesis .