Test the claim that the mean GPA of night students is smaller than 2.1 at the .025 significance level. The null and alternative hypothesis would be: Ho:p = 0.525 Ho: µ = 2.1 Họ:p = 0.525 Ho:µ = 2.1 H9:p= 0.525 Họ:µ= 2.1 H1:p > 0.525 Hị:4 > 2.1 H1:p + 0.525 Hị:µ# 2.1 H:p< 0.525 H1:µ < 2.1 The test is: left-tailed two-tailed right-tailed Based on a sample of 40 people, the sample mean GPA was 2.08 with a standard deviation of 0.02 |(to 2 decimals) The test statistic is: The critical value is: (to 2 decimals) Based on this we: Reject the null hypothesis O Fail to reject the null hypothesis

Test the claim that the mean GPA of night students is smaller than 2.4 at the .05 significance level.The null and alternative hypothesis would be: H0:μ=2.4H0:μ=2.4H1:μ>2.4H1:μ>2.4 H0:p=0.6H0:p=0.6H1:p≠0.6H1:p≠0.6 H0:μ=2.4H0:μ=2.4H1:μ<2.4H1:μ<2.4 H0:p=0.6H0:p=0.6H1:p<0.6H1:p<0.6 H0:p=0.6H0:p=0.6H1:p>0.6H1:p>0.6 H0:μ=2.4H0:μ=2.4H1:μ≠2.4H1:μ≠2.4 The test is: right-tailed two-tailed left-tailed Based on a sample of 60 people, the sample mean GPA was 2.37 with a standard deviation of 0.02The test statistic is:  (to 2 decimals)The critical value is:  (to 2 decimals)

You wish to test the following claim (Ha) at a significance level of a = 0.002. H.: µ1 < µ2 You obtain the following two samples of data. H.:µ1 = µ2 93.9 83.5 80.5 103.9 56.1 81.5 95.2 52 75.1 88 63.1 72 81.5 95.2 84.6 66.7 91.4 100.5 84 62 46 61.7 32.7 118.1 58.6 95.7 95.7 65.6 74.7 61.7 84.6 75.2 73.7 56.6 106.2 84 103.1 84.6 98.3 47.8 50.9 80 62.6 76.4 85.1 101 105.5 91.3 92.2 94.8 78.6 62.6 71.6 105.5 75.2 65.6 Sample #1 Sample #2 54.4 69.3 46.1 96.8 88.3 106.9 106.9 123.5 58.6 55.9 32.6 82 81 81 101 69 63.6 60.4 57.3 86.8 66.7 118.1 82.5 101 66.6 93 62.6 57.9 71.1 69.5 94.5 81.5 93.2 123.5 67.9 94.5 100.1 52 66.7 53.1 103.1 114.8 83.5 76.6 60.3 54.2 What is the test statistic for this sample? (Report answer accurate to three decimal places.) test statistic = What is the p-value for this sample? For this calculation, use the degrees of freedom reported from the technology you are using. (Report answer accurate to four decimal places.) p-value =

Test the claim that the proportion of people who own cats is smaller than 80% at the 0.025 significance level. The null and alternative hypothesis would be: Ho:p = 0.8 Họ:µ > 0.8 Ho:p 0.8 Hị: µ > 0.8 H1:µ + 0.8 H1:p< 0.8 The test is: right-tailed two-tailed left-tailed Based on a sample of 500 people, 75% owned cats The p-value is: (to 2 decimals)

Physical properties of six flame-retardant fabric samples were investigated in an article. Use the accompanying data and a 0.05 significance level to determine whether a linear relationship exists between stiffness x (mg-cm) and thickness y (mm). State the appropriate null and alternative hypotheses. O Ho: P = 0 H₂: P 0 O Ho: P #O H₂: p=0 O Ho: P = 0 H₁₂ :P #0 8.07 24.69 12.33 6.85 24.22 35.69 y 0.29 0.65 0.31 0.27 0.80 0.57 Compute the value of the sample correlation coefficient, r. Round your answer to four decimal places. r= Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to three decimal places.) t = P-value = State the conclusion in the problem context. O Fail to reject Ho. The data does not indicate that the population correlation coefficient differs from 0. Reject Ho. The data does not indicate that the population correlation coefficient differs from 0. Fail to reject Ho. The data indicates that the…

Test the claim that the proportion of men who own cats is smaller than the proportion of women who own cats at the .05 significance level.The null and alternative hypothesis would be:   H0:μM=μFH1:μM<μF   H0:pM=pFH1:pM<pF   H0:pM=pFH1:pM>pF   H0:pM=pFH1:pM≠pF   H0:μM=μFH1:μM≠μF   H0:μM=μFH1:μM>μF The test is: right-tailed left-tailed   two-tailed Based on a sample of 20 men, 40% owned catsBased on a sample of 60 women, 55% owned catsThe test statistic is:  (to 2 decimals)The p-value is:  (to 2 decimals)Based on this we: Fail to reject the null hypothesis Reject the null hypothesis

With 0.05 significance level, H1: p > 0.5, and the value of test statistics is z = 1.25. Then the P-value is equal to 0.8944 -0.1056 0.1056 None

Answer 5-8 based on the table below. Yes No Teachers 12 20 Students 45 23 5) P(teacher | yes) 6) P(no | teacher) 7) P(student | no) 8) P(yes | student)

Test the claim that the mean GPA of night students is significantly different than 3.4 at the 0.2 significance level.The null and alternative hypotheses would be: H0:μ=3.4H0:μ=3.4H1:μ>3.4H1:μ>3.4 H0:p=0.85H0:p=0.85H1:p<0.85H1:p<0.85 H0:p=0.85H0:p=0.85H1:p≠0.85H1:p≠0.85 H0:p=0.85H0:p=0.85H1:p>0.85H1:p>0.85 H0:μ=3.4H0:μ=3.4H1:μ<3.4H1:μ<3.4 H0:μ=3.4H0:μ=3.4H1:μ≠3.4H1:μ≠3.4 Based on a sample of 31 people, the sample mean GPA was 3.37 with a standard deviation of 0.08The p-value is:  (to 3 decimals)The positive significance level is:  (to 2 decimals)

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