A social worker was interested in determining whether there is a significant difference in the average monthly cost per child for childcare outside the home between state supported facilities and privately owned facilities. Two independent random samples were selected yielding the following information.   State supported Facilities Privately owned Facilities Sample Size 64 64 Sample Mean 725 675 Sample Standard Deviation 95 80 Based on the 90% confidence interval for the true mean difference in average cost of children (state supported - privately owned), can one conclude that there is a difference in the average cost of children between the state supported facilities and privately owned facilities?   Group of answer choices A. 90% CI is (-7.28, 52.72). One can conclude that there is a difference in the average cost of childcare between the state supported facilities and privately owned facilities. B. 90% CI is (-7.28, 52.72). One can conclude that there is no difference in the average cost of childcare between the state supported facilities and privately owned facilities. C. 90% CI is (24.46, 75.54). One can conclude that there is no difference in the average cost of childcare between the state supported facilities and privately owned facilities. D. 90% CI is (24.46, 75.54). One can conclude that there is a difference in the average cost of childcare between the state supported facilities and privately owned facilities.

In a test of the effectiveness of garlic for lowering cholesterol, 47 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 3.7 and a standard deviation of 18.8. 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? 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? mg/dL. <μ< mg/dL (Round to two decimal places as needed.). What does the confidence interval suggest about the effectiveness of the treatment? ... OA. The confidence interval limits contain 0, suggesting…

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.

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

Is there any difference in the variability in golf scores for players on the LPGA Tour (the women's professional golf tour) and players on the PGA Tour (the men's professional golf tour)? A sample of 20 tournament scores from LPGA events showed a standard deviation of 2.4623 strokes, and a sample of 30 tournament scores from PGA events showed a standard deviation of 2.2118. Conduct a hypothesis test for equal population variances to determine whether there is any statistically significant difference in the variability of golf scores for male and female professional golfers. Use a = = 0.10. Ho : o Ha : o - Select your answer - - Select your answer What is your conclusion? There is - Select your answer V significant difference in the variances. - Select your answer -

Two groups of individuals were compared with respect to a high carbohydrate low- fat diet (LF) and a high-fat low carbohydrate diet (LC). Mood was assessed using a total mood disturbance score where a lower score is associated with a less negative mood. Assuming that the random samples come from independent normal distributions with a common but unknown variance, construct a 95% confidence interval for the difference between the two population means, LC less LF. Group LC LF 28+22.37 28+18.56 28+22.30 28±8.96 n 12 13 mean 47.3 19.3 standard deviation 28.3 25.8

According to a report on sleep deprivation by the Centers for Disease Control and Prevention, the proportion of California residents who reported insufficient rest or sleep during each of the preceding 30 days is 8.0%, while this proportion is 8.8% for Oregon residents. These data are based on simple random samples of 11,545 California and 4,691 Oregon residents. Calculate a 95% confidence interval for the difference between the proportions of Californians and Oregonians who are sleep deprived and interpret it in context of the data. Also, conduct a hypothesis, with alpha = 0.05, test to determine if these data provide strong evidence the rate of sleep deprivation is different for the two states.     Construct a 95% confidence interval  Which of the following correctly displays the standard error?   0.0046 0.009 0.001 0.00002 0.008