A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.74 hours, with a standard deviation of 2.41 hours. A random sample of 40 adults with children under the age of 18 results in a mean daily leisure time of 4.37 hours, with a standard deviation of 1.92 hours. Construct and interpret a 90% confidence interval for the mean difference in leisure time between adults with no children and adults with children (41 - 42)- Let u, represent the mean leisure hours of adults with no children under the age of 18 and µz represent the mean leisure hours of adults with children under the age of 18. The 90% confidence interval for (H -H2) 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 5.06 hours, with a standard deviation of 2.36 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 1.97 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 μ₁ represent the mean leisure hours of adults with no children under the age of 18 and µ2 represent the mean leisure hours of adults with children under the age of 18. from - The 95% confidence interval for (μ₁ −μ₂) is the range (Round to two decimal places as needed.) hours to hours.

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 4.7 and a standard deviation of 16.2. Construct a 90% 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 µ? Omg/dL <µ< mg/dL (Round to two decimal places as needed.)

The most recent American Time Use Survey found that many Americans barely spend any time reading for fun. People ages 15 to 19 average only 7.8 minutes of leisurely reading per day with a standard deviation of 5.4 minutes. However, people ages 75 and over read for an average of 43.8 minutes per day with a standard deviation of 35.5 minutes. A 95% confidence interval constructed for the true difference in mean amount of time in minutes spent reading per day (15 to 19 - 75 and older) was −36±2.1762−36±2.1762. Does the interval provide convincing evidence that those 75 and older read more than those 15 to 19?