Test the claim that the proportion of men who own cats is smaller than the proportion of women who own cats at the .005 significance level. The null and alternative hypothesis would be: H0:μM=μFH0:μM=μF H1:μM>μFH1:μM>μF   H0:pM=pFH0:pM=pF H1:pMpFH1:pM>pF   H0:pM=pFH0:pM=pF H1:pM≠pFH1:pM≠pF The test is: left-tailed two-tailed right-tailed Based on a sample of 20 men, 30% owned cats Based on a sample of 60 women, 50% owned cats The test statistic is:  (to 2 decimals) The critical value is:  (to 2 decimals) Based on this we: Fail to reject the null hypothesis Reject the null hypothesis

Test the claim that the mean GPA of night students is significantly different than the mean GPA of day students at the 0.02 significance level. (a)  The null and alternative hypothesis would be:  H0:πN=πD H1:πN<πD   H0:μN=μD H1:μN<μD   H0:μN=μD H1:μN>μD   H0:πN=πD H1:πN>πD   H0:πN=πD H1:πN≠πD   H0:μN=μD H1:μN≠μD     (b)  The test is:  right-tailed left-tailed two-tailed   The sample consisted of 45 night students, with a sample mean GPA of 3.02 and a standard deviation of 0.08, and 45 day students, with a sample mean GPA of 3.01 and a standard deviation of 0.03. (c)  The t-score is: (Round to 2 decimal places.) (d)  The p-value is: (Round to 4 decimal places.) (e)  Based on this we conclude: null is plausible strong support for the alternative

Test the claim that the proportion of people who own cats is significantly different than 80% at the 0.1 significance level.The null and alternative hypothesis would be:   H0:μ=0.8H0:μ=0.8H1:μ≠0.8H1:μ≠0.8 H0:p=0.8H0:p=0.8H1:p≠0.8H1:p≠0.8 H0:p≤0.8H0:p≤0.8H1:p>0.8H1:p>0.8 H0:p≥0.8H0:p≥0.8H1:p<0.8H1:p<0.8 H0:μ≤0.8H0:μ≤0.8H1:μ>0.8H1:μ>0.8 H0:μ≥0.8H0:μ≥0.8H1:μ<0.8H1:μ<0.8 The test is:   right-tailed left-tailed two-tailed Based on a sample of 200 people, 87% owned catsThe p-value is:  (to 2 decimals)Based on this we: Reject the null hypothesis Fail to reject the null hypothesis

Test the claim that the mean GPA of night students is larger than the mean GPA of day students at the 0.10 significance level.The null and alternative hypothesis would be:   H0:μN=μDH1:μN≠μD H0:pN≥pDH1:pN<pD H0:μN≤μDH1:μN>μD H0:pN≤pDH1:pN>pD H0:μN≥μDH1:μN<μD H0:pN=pDH1:pN≠pD The test is:   two-tailed right-tailed left-tailed The sample consisted of 80 night students, with a sample mean GPA of 3.36 and a standard deviation of 0.08, and 80 day students, with a sample mean GPA of 3.35 and a standard deviation of 0.04.The test statistic is:  (to 2 decimals)The p-value is:  (to 2 decimals)Based on this we: Reject the null hypothesis Fail to reject the null hypothesis