A study was done using a treatment group and a placebo group. The results are shown in the table. Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. Complete parts (a) and (b) below. Use a 0.01 significance level for both parts. a. Test the claim that the two samples are from populations with the same mean. What are the null and alternative hypotheses? OA. Ho H1 H2 H₁: H1 H2 The test statistic, t, is -1.55. (Round to two decimal places as needed.) The P-value is (Round to three decimal places as needed.) OB. Ho: H1 H2 H₁₁₂ D. Ho: H1 H2 H₁: H1 H2 Treatment Placebo μ H₁ H2 n 25 40 X 2.38 2.65 S 0.53 0.87
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- Astudy was done on body temperatures of men and women. The results are shown in the table. Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. Complete parts (a) and (b) below. Use a 0.05 significance level for both parts.A plant pathologist wants to compare two strains of bacteria in terms of its reproduction rate. Test of normality showed that the two samples have heterogeneous variances. Which of the following would you advise the plant pathologist to conduct? A. t-test B. Mann Whitney C. Kruskal Wallis test . D Anova F-testData on the weights (lb) of the contents of cans of diet soda versus the contents of cans of the regular version of the soda is summarized to the right. Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. Complete parts (a) and (b) below. Use a 0.01 significance level for both parts. a. Test the claim that the contents of cans of diet soda have weights with a mean that is less than the mean for the regular soda. What are the null and alternative hypotheses? OA. Ho: H₁ H₂ H₁: Hy #4₂ OC, Hoi ky tuy H₁: Hy O L P H command n X S Time Remaining: 01:13:11 V : • Diet H₁ 30 0.79861 lb 0.00445 lb ; x { [ option ? I Regular H₂ 30 0.80936 lb 0.00742 lb Next delete
- A study was done using a treatment group and a placebo group. The results are shown in the table. Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. Complete parts (a) and (b) below. Use a 0.05 significance level for both parts. Treatment Placebo μ μ1 μ2 n 27 39 x 2.38 2.65 s 0.87 0.61 a. Test the claim that the two samples are from populations with the same mean. What are the null and alternative hypotheses? A. H0: μ1≠μ2 H1: μ1<μ2 B. H0: μ1<μ2 H1: μ1≥μ2 C. H0: μ1=μ2 H1: μ1>μ2 D. H0: μ1=μ2 H1: μ1≠μ2 Your answer is correct. The test statistic, t, is (Round to two decimal places as needed.)A study was done using a treatment group and a placebo group. The results are shown in the table. Assume that the two samples are independent simple random H samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. Complete parts (a) and (b) below. Use a 0.01 significance level for both parts. a. Test the claim that the two samples are from populations with the same mean. What are the null and alternative hypotheses? OA. Ho: #₁ = 1₂ H₁: H₁ H₂ OC. Ho: H₁A study was done using a treatment group and a placebo group. The results are shown in the table. Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. Complete parts (a) and (b) below. Use a 0.10 significance level for both parts. a. Test the claim that the two samples are from populations with the same mean. What are the null and alternative hypotheses? OA. Ho: H₁ H₂ H₁: Hq ZH₂ OC. Ho: H₁ H₂ H₁: Hy > H₂ The test statistic, t, is. (Round to two decimal places as needed.) (Round to three decimal places as needed.) The P-value is State the conclusion for the test. C... OB. Ho: H₁ H₂ H₁: Hy #H₂ OD. Ho: Hg #U2 H₁: HyChoose the appropriate statistical test. When computing, be sure to round each answer as indicated. A dentist wonders if depression affects ratings of tooth pain. In the general population, using a scale of 1-10 with higher values indicating more pain, the average pain rating for patients with toothaches is 6.8. A sample of 30 patients that show high levels of depression have an average pain rating of 7.1 (variance 0.8). What should the dentist determine? 1. Calculate the estimated standard error. (round to 3 decimals). [st.error] 2. What is thet-obtained? (round to 3 decimals). 3. What is the t-cv? (exact value) 4. What is your conclusion? Only type "Reject" or Retain"Consider the following data: −10, −10, 0, 3, −10, 3, 3 Step 1 of 3: Calculate the value of the sample Variance. Round your answer to one decimal place. Step 2 of 3: Calculate the value of the sample deviation. Round your answer to one decimal place. Step 3 of 3: Calculate the value of the range.Calculate the test statistic (t) and p-value.A study was done using a treatment group and a placebo group. The results are shown in the table. Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. Complete parts (a) and (b) below. Use a 0.01 significance level for both parts. a. Test the claim that the two samples are from populations with the same mean. What are the null and alternative hypotheses? OA. Ho: H₁ H₂ H₁: H₁ H₂ OC. Ho: H₁ H¹/₂ H₁: H₁A firm applies a skill test to applicants to fill the vacant Quality Engineer position. The company's Human Resources Manager knows that the talent exam scores of candidates applying for this position have been distributed normally with 35.8 standard deviations in the past years. A sample of “n” was taken from the candidates who applied to the position of Quality Engineer and aptitude test of these candidates The mean score was 167.5. Find and interpret the 90% confidence interval for the average of the talent exam scores of the applicants who apply for the Quality Engineering position. . (n = the last two digits of your student number. If the last two digits are zero, choose as a two-digit number)Student ID : 201747201Please do 3 and 4.SEE MORE QUESTIONSRecommended textbooks for youMATLAB: An Introduction with ApplicationsStatisticsISBN:9781119256830Author:Amos GilatPublisher:John Wiley & Sons IncProbability and Statistics for Engineering and th…StatisticsISBN:9781305251809Author:Jay L. 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