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 :\(\boxed{\text{Final Answer: The 95% confidence interval for the difference between the mean daily calorie consumption of females in September-February and the mean daily calorie consumption of females in March-August is approximately (-71.97, 6.37). This means we are 95% confident that the true difference in mean daily calorie consumption between the two periods lies within this interval.}}\)