(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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- 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 •H₂Insurance 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.
- 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 12people who buy insurance from Company A, the mean cost is $153 per month with a standard deviation of $16. For 15 randomly selected customers of Company B, you find that they pay a mean of $160 per month with a standard deviation of $10. 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 1 of 3: State the null and alternative hypotheses for the test. Fill in the blank below. H0: μ1=μ2 Ha: μ1_____μ2 Step 2 of 3: Compute the value of the test statistic. Round your answer to three decimal places Step 3 of…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 =light examined data on employment and answered questions regarding why workers separate from their employes. According to the article, the standard deviation of the length of time that women with one job are employed during the first 8 years of their career is 92 weeks. Length of time employed during the first 8 years of career is a left skewed variable. For that variable, do the following tasks. A. determine the sampling distribution of the sample mean for simple random samples of 50 women with one job. Explain your reasoning B. Obtain the probability that the sampling error made in estimating the mean length of time employed by all women with one job by that of a random sample of 50 such women will be at most 20 weeks
- 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.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.
- 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…The salaries of professional baseball players are heavily skewed right with a mean of $3.2 million and a standard deviation of $2 million. The salaries of professional football players are also heavily skewed right with a mean of $1.9 million and a standard deviation of $1.5 million. A random sample of 40 baseball players’ salaries and 35 football players’ salaries is selected. The mean salary is determined for both samples. Let represent the difference in the mean salaries for baseball and football players. Which of the following represents the shape of the sampling distribution for ? skewed right since the populations are both right skewed skewed right since the differences in salaries cannot be negative approximately Normal since both sample sizes are greater than 30 approximately Normal since the sum of the sample sizes is greater than 30
