Using ANOVA, a null hypothesis could look like this: Group of answer choices H0: µ1 =µ2 =µ3 H0: µ1 > µ2 > µ3 H0: µ1 = µ2 > µ3 H0: µ1 ≠ µ2 ≠ µ3
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Explanation:Here given the multiple choices and find the null hypothesis using Anova
So here null hypothesis could look like this is option A
Clearly in step2
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- 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 alternativeTest the claim that the mean GPA of night students is larger than 3.2 at the .005 significance level.The null and alternative hypothesis would be: H0:p=0.8H1:p<0.8 H0:p=0.8H1:p>0.8 H0:μ=3.2H1:μ>3.2 H0:μ=3.2H1:μ≠3.2 H0:μ=3.2H1:μ<3.2 H0:p=0.8H1:p≠0.8 Based on a sample of 55 people, the sample mean GPA was 3.22 with a standard deviation of 0.03The test statistic is (to 3 decimals)The critical value is (to 3 decimals)Based on this we fail to reject the null hypothesis reject the null hypothesisYou wish to test the following claim (HaHa) at a significance level of α=0.01α=0.01. Ho:μ1=μ2Ho:μ1=μ2 Ha:μ1≠μ2Ha:μ1≠μ2You obtain the following two samples of data. Sample #1 Sample #2 65.8 77.7 105.1 78.6 89.2 69.1 96.7 91.3 105.9 87.5 70.5 93.9 83.4 88.3 79.5 66.7 83.4 91.7 106.7 73 75.7 98.7 76.7 80 53.2 92.1 93 78.2 72.4 79.1 82.1 74.1 78.2 85.5 85.9 99.3 84.2 104.3 88.3 107.6 63.8 99.8 97.7 86.3 74.9 72.7 77.5 93.9 76.1 77.1 81.9 74.1 73.6 78.7 81.9 78.9 81.1 88 85.8 75.3 74.3 77.5 69.1 75.5 72.9 80.1 84.3 73.1 67.3 72.2 76.1 67.8 74.9 67.3 74.7 70.1 65.4 76.9 83.8 79.7 72.9 81.7 74.1 68.2 65.4 80.9 69.4 79.5 96 77.5 77.1 75.9 68.7 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…
- Test the claim that the proportion of men who own cats is smaller than 90% at the .05 significance level.The null and alternative hypothesis would be: H0:μ=0.9H0:μ=0.9H1:μ>0.9H1:μ>0.9 H0:p=0.9H0:p=0.9H1:p>0.9H1:p>0.9 H0:p=0.9H0:p=0.9H1:p≠0.9H1:p≠0.9 H0:μ=0.9H0:μ=0.9H1:μ≠0.9H1:μ≠0.9 H0:μ=0.9H0:μ=0.9H1:μ<0.9H1:μ<0.9 H0:p=0.9H0:p=0.9H1:p<0.9H1:p<0.9 The test is: right-tailed two-tailed left-tailed Based on a sample of 65 people, 87% owned catsThe 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 I dont understand this.Test the claim that the proportion of men who own cats is significantly different than the proportion of women who own cats at the 0.2 significance level.The null and alternative hypothesis would be: H0:pM=pFH0:pM=pFH1:pM<pFH1:pM<pF H0:μM=μFH0:μM=μFH1:μM≠μFH1:μM≠μF H0:pM=pFH0:pM=pFH1:pM≠pFH1:pM≠pF H0:pM=pFH0:pM=pFH1:pM>pFH1:pM>pF H0:μM=μFH0:μM=μFH1:μM>μFH1:μM>μF H0:μM=μFH0:μM=μFH1:μM<μFH1:μM<μF The test is: two-tailed right-tailed left-tailed Based on a sample of 60 men, 30% owned catsBased on a sample of 40 women, 40% owned catsThe 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 Check AnswerQuestion 14If α = .05 for a two-tailed hypothesis test, how are the boundaries for the critical region determined? a. Boundaries are drawn so there is 10% (.10) in each tail of the distribution. b. Boundaries are drawn so there is 5% (.05) in each tail of the distribution. c. Boundaries are drawn so there is 5% (.05) in the center of the distribution. d. Boundaries are drawn so there is 2.5% (.025) in each tail of the distribution.
- You wish to test the following claim (HaHa) at a significance level of α=0.001α=0.001. Ho:μ1=μ2Ho:μ1=μ2 Ha:μ1≠μ2Ha:μ1≠μ2You obtain the following two samples of data. Sample #1 Sample #2 50.8 84.4 72.1 78.4 72.9 48.8 87.4 36.1 78.8 51.7 74.7 56.3 65.2 98.6 72.5 72.1 72.9 54.2 69.8 73.4 51.7 68.5 59.4 86.8 48.8 47.6 48.8 85.6 65.2 43.4 76.1 69.8 43.4 69.4 55.6 74.3 108.9 92.4 77.9 46.4 47.6 93.3 60 72.9 74.7 66.6 72 84.3 66.1 75 86.5 96.6 47.1 79.1 75 74.5 82.2 77.8 79.5 70.9 71.5 104.8 88.8 81.7 99.8 89.3 91.2 87.9 111.1 66.1 68 79.5 82.6 61.4 90.7 115.1 106.9 60.4 50.1 85.7 61.4 100.5 105.8 79.1 88.3 73 68.6 57.1 70.9 67.4 42.4 52.3 80.4 88.8 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…Test the claim that the proportion of people who own cats is significantly different than 80% at the 0.2 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: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 H0:p=0.8H0:p=0.8H1:p≠0.8H1:p≠0.8 The test is: two-tailed right-tailed left-tailed Based on a sample of 300 people, 74% 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 hypothesisTest the claim that the proportion of men who own cats is significantly different than the proportion of women who own cats at the 0.2 significance level.The null and alternative hypothesis would be: H0:μM=μFH0:μM=μFH1:μM<μFH1:μM<μF H0:pM=pFH0:pM=pFH1:pM<pFH1:pM<pF H0:pM=pFH0:pM=pFH1:pM≠pFH1:pM≠pF H0:pM=pFH0:pM=pFH1:pM>pFH1:pM>pF H0:μM=μFH0:μM=μFH1:μM>μFH1:μM>μF H0:μM=μFH0:μM=μFH1:μM≠μFH1:μM≠μF The test is: right-tailed left-tailed two-tailed Based on a sample of 40 men, 25% owned catsBased on a sample of 20 women, 30% owned catsThe 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
- You are conducting a study to see if the proportion of men over 50 who regularly have their prostate examined is significantly more than 0.69. You use a significance level of α=0.01α=0.01. H0:p=0.69H0:p=0.69 H1:p>0.69H1:p>0.69You obtain a sample of size n=445n=445 in which there are 338 successes.What is the p-value for this sample? (Report answer accurate to four decimal places.)p-value = The p-value is... less than (or equal to) αα greater than αα This p-value leads to a decision to... reject the null accept the null fail to reject the null As such, the final conclusion is that... There is sufficient evidence to warrant rejection of the claim that the proportion of men over 50 who regularly have their prostate examined is more than 0.69. There is not sufficient evidence to warrant rejection of the claim that the proportion of men over 50 who regularly have their prostate examined is more than 0.69. The sample data support the claim that the proportion of men…(see attachment) Use technology to conduct a one-factor ANOVA to determine if the group means are equal using α=0.05α=0.05. Group means (report to 2 decimal places): Group 1: Internship: Group 2: Co-op: Group 3: Work Study: ANOVA summary statistics:F-ratio = (report accurate to 3 decimal places)p=p= (report accurate to 4 decimal places)You wish to test the following daim (Ha) at a significance level of a = 0.005. H.:P1 > P2 Ha:P1 < P2 You obtain a sample from the first population with 303 successes and 238 failures. You obtain a sample from the second population with 155 successes and 79 failures. critical value = [three decimal accuracy] test statistic = [three decimal accuracy] The test statistic is... O in the critical region O not in the critical region This test statistic leads to a decision to... reject the null hypothesis fail to reject the null hypothesis As such, the final condusion is that... O There is sufficient evidence to support that the first population proportion is less than the second population proportion. O There is not sufficient evidence to support that the first population proportion is less than the second population proportion.