For a new study conducted by a fitness magazine, 245 females were randomly selected. For each, the mean daily calorie consumption was calculated for a September-February period. A second sample of 240 females was chosen independently of the first. For each of them, the mean daily calorie consumption wa calculated for a March-August period. During the September-February period, participants consumed a mean of 2383.6 calories daily with a standard deviatio of 190. During the March-August period, participants consumed a mean of 2416.4 calories daily with a standard deviation of 245 . 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 $95 \%$ confidence interval for $\mu_{1}-\mu_{2}$, the difference between the mean daily calorie consumption $\mu_{1}$ of females in September-February and the mean daily calorie consumption $\mu_{2}$ of females in March-August. Then find the lower limit and upper limit of the $95 \%$ confidence interval.
Carry your intermediate computations to at least three decimal places. Round your answers to at least two decimal places. (If necessary, consult a list of formulas.)
Answer
\(\boxed{\text{Final Answer: The 95% confidence interval for the difference between the mean daily calorie consumption of females in September-February and the mean daily calorie consumption of females in March-August is approximately (-71.97, 6.37). This means we are 95% confident that the true difference in mean daily calorie consumption between the two periods lies within this interval.}}\)
Step 1 ：Given values are: sample mean of daily calorie consumption from September to February (\(x_1\)) = 2383.6, standard deviation (\(s_1\)) = 190, sample size (\(n_1\)) = 245, sample mean of daily calorie consumption from March to August (\(x_2\)) = 2416.4, standard deviation (\(s_2\)) = 245, sample size (\(n_2\)) = 240, and significance level (\(\alpha\)) = 0.05.
Step 2 ：Calculate the difference of means (\(diff\)) = \(x_1 - x_2\) = -32.80000000000018.
Step 3 ：Calculate the standard error (\(se\)) = \(\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}\) = 19.93617579783487.
Step 4 ：Calculate the degrees of freedom (\(df\)) = \(n_1 + n_2 - 2\) = 483.
Step 5 ：Find the t-score for \(\alpha/2\) = 1.964887640611612.
Step 6 ：Calculate the confidence interval. The lower limit = \(diff - t_{score} \times se\) = -71.97234542622627 and the upper limit = \(diff + t_{score} \times se\) = 6.372345426225898.
Step 7 ：\(\boxed{\text{Final Answer: The 95% confidence interval for the difference between the mean daily calorie consumption of females in September-February and the mean daily calorie consumption of females in March-August is approximately (-71.97, 6.37). This means we are 95% confident that the true difference in mean daily calorie consumption between the two periods lies within this interval.}}\)
