A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.77 hours, with a standard deviation of 2.34 hours. A random sample of 40 adults with children under the age of 18 results in a mean daily leisure time of 4.06 hours, with a standard deviation of 1.56 hours. Construct and interpret a 95% confidence interval for the mean difference in leisure time between adults with no children and adults with children (H1 - H2). Let u, represent the mean leisure hours of adults with no children under the age of 18 and H, represent the mean leisure hours of adults with children under the age of 18. The 95% confidence interval for (H1 - H2) is the range from hours to hours. (Round to two decimal places as needed.) What is the interpretation of this confidence interval? O A. There is a 95% probability that the difference of the means is in the interval. Conclude that there is insufficient evidence of a significant difference in the number of leisure hours. O B. There is 95% confidence that the difference of the means is in the interval. Conclude that there is a significant difference in the number of leisure hours. OC. There is a 95% probability that the difference of the means is in the interval. Conclude that there is a significant difference in the number of leisure hours. O D. There is 95% confidence that the difference of the means is in the interval. Conclude that there is insufficient evidence of a significant difference in the numbor of loicuro houre

A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.17 hours, with a standard deviation of 2.45 hours. A random sample of 40 adults with children under the age of 18 results in a mean daily leisure time of 4.44 hours, with a standard deviation of 1.71 hours. Construct and interpret a 95% confidence interval for the mean difference in leisure time between adults with no children and adults with children (u1-2) Let u, represent the mean leisure hours of adults with no children under the age of 18 and u, represent the mean leisure hours of adults with children under the age of 18. The 95% confidence interval for (H, -µ2) is the range from hours to hours. (Round to two decimal places as needed.)

A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.88 hours, with a standard deviation of 2.49 hours. A random sample of 40 adults with children under the age of 18 results in a mean daily leisure time of 4.29 hours, with a standard deviation of 1.76 hours. Construct and interpret a 90% confidence interval for the mean difference in leisure time between adults with no children and adults with children (u, - H2). Let u, represent the mean leisure hours of adults with no children under the age of 18 and µ, represent the mean leisure hours of adults with children under the age of 18. The 90% confidence interval for (µ1 - H2) is the range from hours to hours. (Round to two decimal places as needed.)

In a test of the effectiveness of garlic for lowering cholesterol, 48 subjects were treated with garlic in a processed tablet form. Cholesterol levels were measured before and after the treatment. The changes (before - after) in their levels of LDL cholesterol (in mg/dL) have a mean of 3.6 and a standard deviation of 17.5. Construct a 95% confidence interval estimate of the mean net change in LDL cholesterol after the garlic treatment. What does the confidence interval suggest about the effectiveness of garlic in reducing LDL cholesterol? Click here to view a t distribution table. Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table. What is the confidence interval estimate of the population mean u? mg/dL <µ< mg/dL (Round to two decimal places as needed.)

A marketing researcher wants to estimate the mean amount spent​ ($) on a certain retail website by members of the​ website's premium program. A random sample of 91 members of the​ website's premium program who recently made a purchase on the website yielded a mean of​$1700 and a standard deviation of ​$200. Construct a 95​% confidence interval estimate for the mean spending for all shoppers who are members of the​ website's premium program.

In a test of the effectiveness of garlic for lowering cholesterol, 42 subjects were treated with garlic in a processed tablet form. Cholesterol levels were measured before and after the treatment. The changes (before - after) in their levels of LDL cholesterol (in mg/dL) have a mean of 2.0 and a standard deviation of 16.5. Construct a 99% confidence interval estimate of the mean net change in LDL cholesterol after the garlic treatment. What does the confidence interval suggest about the effectiveness of garlic in reducing LDL cholesterol?

Bone mineral density (BMD) is a measure of bone strength. Studies show that BMD declines after age 45. The impact of exercise may increase BMD. A random sample of 59 women between the ages of 41 and 45 with no major health problems were studied. The women were classified into one of two groups based upon their level of exercise activity: walking women and sedentary women. The 39 women who walked regularly had a mean BMD of 5.96 with a standard deviation of 1.22. The 20 women who are sedentary had a mean BMD of 4.41 with a standard deviation of 1.02. Which of the following inference procedures could be used to estimate the difference in the mean BMD for these two types of women

Based on a random sample of 1,120 Americans 15 years of age or older, the mean amount of time spent eating or drinking each day is 1.23 hours. Assuming that the population standard deviation for amount of time eating or drinking is 0.65 hour, construct and interpret a 90% confidence interval for the mean amount of time spent eating or drinking each day by Americans age 15 years or older.

A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.75 hours, with a standard deviation of 2.47 hours. A random sample of 40 adults with children under the age of 18 results in a mean daily leisure time of 4.03 hours, with a standard deviation of 1.78 hours. Construct and interpret a 95% confidence interval for the mean difference in leisure time between adults with no children and adults with children (u, - H2). Let H, represent the mean leisure hours of adults with no children under the age of 18 and u, represent the mean leisure hours of adults with children under the age of 18. The 95% confidence interval for (µ1 - H2) is the range from hours to hours. (Round to two decimal places as needed.)