Name:Show work when necessary. Each part is worth the same amount. Assume that all necessary conditions are methypothesis tests we have studied.1) Are low-fat diets or low-carb diets more effective for weight loss? A sample of 78 subjects went ona low-carbohydrate diet for six months. At the end of that time, the sample mean weight loss was15.0 pounds with a sample standard deviation of 7.77 pounds. A second sample of 74 subjects wenton a low-fat diet. Their sample mean weight loss was 16.3 with a standard deviation of 6.10. Canyou conclude that the mean weight loss differed between the two diets? Use the a = 0.01 level.i). State the appropriate null and alternate hypotheses.ii). Compute the test statistic.iii). How many degrees of freedom are there, using the simple method?iv). Do you reject Ho? State a conclusion.

There are two independent samples which are low-carbohydrate diet and low-fat diet. We have to test…

Answered: 1) Are low-fat diets or low-carb diets…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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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.
Insurance companies track life expectancy information to assist in determining the cost of life insurance policies. Last year the average life expectancy of all policyholders was 77 years. ABI Insurance wants to determine if their clients now have a longer life expectancy, on average, so they randomly sample 20 of their recently paid policies. The sample has a mean of 78.6 years and a standard deviation of 4.48 years. The p-value associated with the resulting test statistic is 0.063. At a = 0.05, which of the following is the correct conclusion? O A. We fail to reject the null hypothesis. O B. We reject the null hypothesis. OC. There is not significant evidence to indicate an increase in average life expectancy. O D. Both A and C O E. Both B and C
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 16.7. Use a 0.01 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?
Two brands of batteries are tested, and their voltages are compared. The summary statistics follow. For sample 1, the sample size is 27; the population standard deviation is 0.3 volts and the mean is 9.2 volts. For sample 2, the sample size is 30; the population standard deviation is 0.1 volts and the mean is 8.8 volts. Is the mean of sample 1 smaller than the mean of sample 2? Using a significance level of 0.05. a) Identify the claim and state the hypotheses. b) What is the test statistics? Find the critical values (s). c) Make the decision to reject or not reject the null hypothesis and explain why. d) Graph your decision and include all the values. That is, the mean, the standard deviation, the critical value(s), the rejection region. e) Set up the formula with the correct numbers for the 95% confidence interval of the mean number of jobs.
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 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
An educator has developed a program to improve math scores on the Texas STAAR test. The average STAAR Mathematics test for the third graders in the district was a mean= 1472. A sample of n= 30 third graders took the new math training prgram before taking the STAAR Math test. The average score for the students was M= 1497. with a standard deviation of s= 145 (note, this is standard deviation, not variance). Was there a significant change in test scores ( use stard devistion of 0.05, two tail test). Calculate the effect size using Cohen's d.
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 15 people who buy insurance from Company A, the mean cost is $154 per month with a standard deviation of $13. For 11 randomly selected customers of Company B, you find that they pay a mean of $159 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 0.02 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.
Dr. Graham is interested in determining if middle-aged adults use text messaging more or less frequently than the general population. Dr. Graham collects information on text messaging from a random sample of 50 adults ages 25 to 44. Dr. Graham finds that these individuals send or receive an average of 68 text messages per day. Using the population mean (and standard deviation) of 41.5 texts per day (34 texts per day), determine whether adults in this age group use text messaging more than the general public.
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.
Beer Drinking. The mean annual consumption of beer per person in the US is 22.0 gallons . A random sample of 300 Washington D.C. residents yielded a mean annual beer consumption of 27.8 gallons. At the 10% significance level, do the data provide sufficient evidence to conclude that the mean annual consumption of beer per person for the nation’s capital differs from the national mean? Assume that the standard deviation of annual beer consumption for Washington D.C. residents is 55 gallons.
In September a biological research team caught, weighed, and released a random sample of 24 chipmunks in Rocky Mountain National Park. The mean of the sample weights was x-bar = 8.7 oz. with sample standard deviation s = 1.4 oz. Find a 90% confidence interval for the mean September weight of all chipmunks in Rocky Mountain National Park.
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…

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