Question

For a new study conducted by a fitness magazine,

240

females were randomly selected. For each, the mean daily calorie consumption was calculated for a September-February period. A second sample of

210

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

2385.5

calories daily with a standard deviation of

222

. During the March-August period, participants consumed a mean of

2414.5

calories daily with a standard deviation of

252.5

. 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

−μ1μ2

, the difference between the mean daily calorie consumption

μ1

of females in September-February and the mean daily calorie consumption

μ2

of females in March-August. Then complete the table below

What is the lower limit of the 90% confidence interval?
What is the upper limit of the 90% confidence interval?

Answer 1

Since we don't know population S.D.s. So, we calculate them using formula:

lower limit = -31.495

upper limit = -26.505