(a) Construct a 99.8% confidence interval for the difference in mean weight loss between the low-carb and low-fat diets. Let M, denote the mean weight loss for low-carb diet. Use the TI-84 calculator and round the answers to one decimal place.
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- Insurance Company A claims that its customers pay less for car insurance, on average, than customers of its competitor, Company B. You wonder if this is true, so you decide to compare the average monthly costs of similar insurance policies from the two companies. For a random sample of 7 people who buy insurance from Company A, the mean cost is $150 per month with a standard deviation of $16. For 12 randomly selected customers of Company B, you find that they pay a mean of $160 per month with a standard deviation of $14. Assume that both populations are approximately normal and that the population variances are equal to test Company A's claim at the 0.10 level of significance. Let customers of Company A be Population 1 and let customers of Company B be Population 2. Step 2 of 3: Compute the value of the test statistic. Round your answer to three decimal places.Low-fat or low-carb? Are low-fat diets or low-carb diets more effective for weight loss? A sample of 77 subjects went on a low-carbohydrate diet for six months. At the end of that time, the sample mean weight loss was 4.7 kilograms with a sample standard deviation of 7.16 kilograms. A second sample of 79 subjects went on a low-fat diet. Their sample mean weight loss was 2.6 kilograms with a standard deviation of 5.90 kilograms. Can you conclude that the mean weight loss differs between the two diets? Let 4, denote the mean weight lost on the low-carb diet and u, denote the mean weight lost on the low-fat diet. Use the a = 0.01 level and the TI-84 Plus calculator. Compute the -value. Round the answer to four decimal places. Determine whether to reject H0 State a conclusionInsurance Company A claims that its customers pay less for car insurance, on average, than customers of its competitor, Company B. You wonder if this is true, so you decide to compare the average monthly costs of similar insurance policies from the two companies. For a random sample of 13 people who buy insurance from Company A, the mean cost is $150 per month with a standard deviation of $19. For 9 randomly selected customers of Company B, you find that they pay a mean of $157 per month with a standard deviation of $16. Assume that both populations are approximately normal and that the population variances are equal to test Company A's claim at the 0.05 level of significance. Let customers of Company A be Population 1 and let customers of Company B be Population 2. Step 1 of 3: State the null and alternative hypotheses for the test. Fill in the blank below. Ho: M₁ = μ₂ Ha:M₁ •H₂
- In a test of the effectiveness of garlic for lowering cholesterol, 81 subjects were treated with raw garlic. Cholesterol levels were measured before and after the treatment. The changes (before minus after) in their levels of LDL cholesterol (in mg/dL) have a mean of 0.9 and a standard deviation of 20.8. Use a 0.05 significance level to test the claim that with garlic treatment, the mean change in LDL cholesterol is greater than 0. What do the results suggest about the effectiveness of the garlic treatment? Assume that a simple random sample has been selected. Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim. .... What are the null and alternative hypotheses? O B. Ho: H= 0 mg/dL O A. Ho: u= 0 mg/dL H,: µ0 mg/dL H1: µ> 0 mg/dL H,: µ<0 mg/dL Determine the test statistic. 0.39 (Round to two decimal places as needed.) Determine the P-value (Round to three decimal places as needed.)Fear of heights - A psychologist has developed a new treatment for acrophobia (the extreme fear of heights) and wants to compare the results of this new treatment to the results of the standard treatment. A study is conducted where a random sample of 16 patients undergo the new treatment and their improvement scores on a diagnostic test are compared to a control sample of 13 patients who received the standard treatment. Summary scores for both groups are shown in the table below. Treatment Sample Mean Sample Standard Deviation New 50.9 12 Standard 47.2 11.4 The degrees of freedom for this problem is d f 26.287709. 1. Select the hypotheses that should be used to assess if the new treatment results in higher average improvement scores than the standard treatment. А. Но : И1 = µ2 vs. Ha : H1 # µ2 OB. Ho : µ1 = µ2 vS. Ha : H1 H2 2. Calculate the test statistic. ? V = 3. Calculate the p-value. p-value =Choose 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"
- Insurance Company A claims that its customers pay less for car insurance, on average, than customers of its competitor, Company B. You wonder if this is true, so you decide to compare the average monthly costs of similar insurance policies from the two companies. For a random sample of 1313 people who buy insurance from Company A, the mean cost is $151$151 per month with a standard deviation of $16$16. For 99 randomly selected customers of Company B, you find that they pay a mean of $158$158 per month with a standard deviation of $19$19. Assume that both populations are approximately normal and that the population variances are equal to test Company A’s claim at the 0.050.05 level of significance. Let customers of Company A be Population 1 and let customers of Company B be Population 2. Step 2 of 3: Compute the value of the test statistic. Round your answer to three decimal places.In a test of the effectiveness of garlic for lowering cholesterol, 49 subjects were treated with raw garlic. Cholesterol levels were measured before and after the treatment. The changes (before minus after) in their levels of LDL cholesterol (in mg/dL) have a mean of 0.6 and a standard deviation of 20.7. Use a 0.05 significance level to test the claim that with garlic treatment, the mean change in LDL cholesterol is greater than 0. What do the results suggest about the effectiveness of the garlic treatment? Assume that a simple random sample has been selected. Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim. What are the null and alternative hypotheses? A. Ho: μ = 0 mg/dL H₁: μ> 0 mg/dL C. Ho: μ = 0 mg/dL H₁: μ#0 mg/dL Determine the test statistic. (Round to two decimal places as needed.) Determine the P-value. (Round to three decimal places as needed.) State the final conclusion that addresses the…Do political science classes require more writing than history classes? The 54 randomly selected political science classes assigned an average of 16.4 pages of essay writing for the course. The standard deviation for these 54 classes was 4.8 pages. The 52 randomly selected history classes assigned an average of 15.8 pages of essay writing for the course. The standard deviation for these 52 classes was 4.6 pages. What can be concluded at the a = 0.10 level of significance? For this study, we should use Select an answer a. The null and alternative hypotheses would be: Ho: Select an answer Select an answer Select an answer H1: Select an answer Select an answer Select an answer b. The test statistic ? V = (please show your answer to 3 decimal places.) c. The p-value = d. The p-value is ? Va e. Based on this, we should Select an answer f. Thus, the final conclusion is that ... |(Please show your answer to 4 decimal places.) ] the null hypothesis. O The results are statistically…
- Do political science classes require less writing than history classes? The 47 randomly selected political science classes assigned an average of 19 pages of essay writing for the course. The standard deviation for these 47 classes was 5.6 pages. The 56 randomly selected history classes assigned an average of 21.9 pages of essay writing for the course. The standard deviation for these 56 classes was 4.1 pages. What can be concluded at the a = 0.10 level of significance? For this study, we should use Select an answer a. The null and alternative hypotheses would be: Ho: Select an answer v Select an answer v Select an answer v H: Select an answer v Select an answer v Select an answer v b. The test statistic ? v = (please show your answer to 3 decimal places.) c. The p-value = (Please show your answer to 4 decimal places.) d. The p-value is ? va e. Based on this, we should Select an answer v the null hypothesis. f. Thus, the final conclusion is that ... O The results are statistically…Low-fat or low-carb? Are low-fat diets more effective for weight loss? A sample of 46 subjects went on a low carbohydrate diet for six months. At the end of that time, the sample mean weight loss was 4.6 kilograms with a sample standard deviation of 6.28 kilograms. A second sample of 49 subjects went on a low-fat diet. Their sample mean weight loss was 2.3 kilograms with a standard deviation of 4.66 kilograms. Can you conclude that the mean weight loss differs between the two diets? Let μ1 denote the mean weight lost on the low-carb diet and μ2 denote the mean weight lost on the low-fat diet. Use the =α0.01level and the P-value method. Compute the test statistic. Round the answer to three decimal places. t =Do political science classes require more writing than history classes? The 40 randomly selected political science classes assigned an average of 13.1 pages of essay writing for the course. The standard deviation for these 40 classes was 3.9 pages. The 53 randomly selected history classes assigned an average of 10.8 pages of essay writing for the course. The standard deviation for these 53 classes was 5.6 pages. What can be concluded at the a = 0.05 level of significance? For this study, we should use Select an answer ↑ a. The null and alternative hypotheses would be: Ho: Select an answer Select an answer Select an answer H₁: Select an answer Select an answer Select an answer b. The test statistic ? (please show your answer to 3 decimal places.) c. The p-value = (Please show your answer to 4 decimal places.) d. The p-value is ? a e. Based on this, we should Select an answer the null hypothesis. f. Thus, the final conclusion is that …..