Question

answer neatly and correctly please!

For a new study conducted by a fitness magazine, 275 females were randomly selected. For each, the mean daily calorie consumption was calculated for a September-February period. A second sample of 250 females was chosen independently of the first. For each of them, the mean daily calorie consumption was calculated for a March-August period. During the September-February period, participants consumed a mean of 2386.1 calories daily with a standard deviation of 230. During the March-August period, participants consumed a mean of 2413.9 calories daily with a standard deviation of 262.5. The population standard deviations of daily calories consumed for females in the two periods can be estimated using the sample standard deviations, as the samples that were used to compute them were quite large. Construct a 90% confidence interval for \1-12, the difference between the mean daily calorie consumption of females in September-February and the mean daily calorie consumption of females in March-August. Then complete the table below. Carry your intermediate computations to at least three decimal places. Round your answers to at least two decimal places. What is the lower limit of the 90% confidence interval? What is the upper limit of the 90% confidence interval? x 6 ?

Answer 1

The statistic software output for this problem is:

sample Z summary confidence interval: 1 Mean of population 1 (Std. dev. = 230) H/2: Mean of population 2 (Std. dev. = 262.5) 1-2 Difference between two means 90% confidence interval results: Difference n1 n2 Sample mean Std.err L. limit U. limit 275 250 -27.8 21.633045 -63.383193 7.7831926

The 90% confidence interval is :

Lower limit = -63.38

Upper limit = 7.78