Raw data in the table below come from an independent-measures experiment comparing three different treatment conditions. Treatment 1 0 1 0 3 X=1 Treatment 2 1 4 1 2 X=2 Should you reject or fail to reject the null hypothesis? Reject Fail to reject Treatment 3 3 3 4 5 X=3.75 To evaluate the effect of the independent variable, researchers conducted the independent-measures one-way ANOVA. They computed the sum of squares between groups to be 15.5 and the sum of squares within groups 14.75.
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- Researchers investigate how the presence of cell phones influence the quality of human interaction. Subjects are randomly selected from a population and divided into an experimental group that is asked to leave their phones in the front of the room and a control group that are not asked to leave their cell phones at the front of the room. Subjects are left alone for 10 minutes and then asked to take a survey designed to measure quality of interactions they had with others in the experiment. What statistical test is appropriate?Using traditional methods it takes 8.8 hours to receive a basic driving license. A new license training method using Computer-Aided Instruction (CAI) has been proposed. A researcher used the technique on 17 students and observed that they had a mean of 8.4 hours with a variance of 2.89. Is there evidence at the 0.1 level that the technique performs differently than the traditional method? Assume the population distribution is approximately normal. Step 4 of 5 : Determine the decision rule for rejecting the null hypothesis. Round your answer to three decimal places.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
- The following data summarize the results from an independent-measures study comparing three treatment conditions. Treatment I II III N = 12 3 5 6 G = 60 5 5 10 ∑X2 = 392 3 1 10 1 5 6 x̅1 = 3 x̅2 = 4 x̅3 = 8 T1 = 12 T2 = 16 T3 =32…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: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:pM=pFH0:pM=pFH1:pM<pFH1:pM<pF H0:μM=μFH0:μM=μFH1:μM<μFH1:μM<μF The test is: left-tailed two-tailed right-tailed Based on a sample of 20 men, 40% owned catsBased on a sample of 40 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 hypothesisAnalysis of Variance (ANOVA) Researchers conducted a study to evaluate the effectiveness of three alcohol treatment programs. The researchers randomly assigned each of the 15 patients to receive one of these treatments. Sixty days later, the researchers measured the effectiveness of the program. One of their measures was the patient’s self-reported need to use alcohol. It was measured on a scale of 1 (low need) to 10 (high need). Test the null hypothesis that the patient’s self-reported need to use alcohol does not differ by type of treatment. Write a concluding statement. Treatment A Treatment B Treatment C 8 6 6 10 5 5 10 10 2 8 2 2 10 4 1
- Test the claim that the proportion of men who own cats is significantly different than 80% at the 0.1 significance level.The null and alternative hypothesis would be: 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 H0:μ=0.8H0:μ=0.8H1:μ≠0.8H1:μ≠0.8 The test is: two-tailed right-tailed left-tailed Based on a sample of 100 people, 87% owned catsThe test statistic is: (to 2 decimals)The positive critical value is: (to 2 decimals)Based on this we: Fail to reject the null hypothesis Reject the null hypothesisA new method was developed to reduce variability of test scores by eliminating lower scores. Two groups, one called the control group and the other, the experimental group, both took the same test. The experimental group was taught using the new method. Do the data provide sufficient evidence to conclude that there is less variation among scores when the new method is used? Perform an F-test at the 1% significance level. (Note: s, = 6.3 and Control Experimental 30 19 29 26 33 36 29 18 33 31 S2 = 3.3.) 33 18 19 32 27 32 29 30 33 ... .. First find the null and alternative hypotheses. Which of the following correctly states the hypotheses? O A. Ho: 01 72 Ha: 01 = 02 O B. Ho: 01 >02 Ha: 0102 Ha: 01 #02 O E. Ho: 01 =02 Ha: 01 02 nts SCO Compute the test statistic F. Scor (Round to three decimal places as needed.) essUsing traditional methods it takes 8.2 hours to receive a basic flying license. A new license training method using Computer Aided Instruction (CAI) has been proposed. A researcher used the technique on 26 students and observed that they had a mean of 8.0 hours with a variance of 2.89. Is there evidence at the 0.1 level that the technique reduces the training time? Assume the population distribution is approximately normal. Step 4 of 5: Determine the decision rule for rejecting the null hypothesis. Round your answer to three decimal places.
- In an industrial experiment, a job was performed by 10 workers using Method I, and by 7 workers using Method II. The experiment yielded the following data. The unit of measurement is minute. A statistician wants to test for the equality of two variances and the difference in the mean times to complete the job using the two methods. (This is a case of two independent samples drawn from two normal populations). Method 1 Method 2 12221915101615222613 61071261012 n1=10 Mean=17 Variance=26 n2=7 Mean=9 Variance=7 (a) Test at the 5% significance level whether the two variances are equal. (Give null and the alternative hypotheses, test statistic, rejection rule, conclusion) (b) Estimate with 95% confidence the ratio of the two population variances. Briefly describe what the interval estimate tells you. (c) Briefly explain how to use the interval estimate in part (b) to test the hypotheses in (a).Considering a treadmill test given to patients being tested for high blood pressure, male patients took their pulse rates before and after running for 5 min. Subject 1 2 3 4 5 6 7 8 9 10 Pulse before 64 100 85 60 92 85 68 85 85 68 Pulse after 68 115 84 68 105 92 72 88 80 92 Using a 0.05 significance level, perform the 8-step hypothesis test to test the claim that the mean difference between the pulse rates before and after the run is significantly zero. Based on the result, do the male pulse rates taken before and after running appear to be about the same or not?