Final answer:
The 95% confidence interval for the mean daily intake of dairy products for men is constructed by calculating the margin of error using a z-score of 1.96 and the sample information provided, resulting in a confidence interval of (746.31, 765.69) grams.
Step-by-step explanation:
To construct a 95% confidence interval for the mean daily intake of dairy products for men, given a sample mean of 756 grams per day and a standard deviation of s = 35 grams per day for a sample size of n = 50, we would use the following steps:
- Firstly, determine the appropriate z-score that corresponds to the middle 95% of the standard normal distribution. Since typical z-tables give the area to the left of the z-value, we look for a z-value such that the area to the left is 0.975 (which includes 47.5% to the left of the mean and 47.5% to the right, giving 95% in total). This z-value is approximately 1.96.
- Next, calculate the standard error of the mean (SEM), which is the standard deviation divided by the square root of the sample size: SEM = s / sqrt(n) = 35 / sqrt(50).
- Then, compute the margin of error (ME) using the z-score and the SEM: ME = z * SEM.
- Finally, construct the confidence interval (CI) using the sample mean and the margin of error: CI = mean ± ME.
Applying these steps, we would have SEM = 35 / sqrt(50) = 4.95 grams and ME = 1.96 * 4.95 ≈ 9.69 grams. The 95% confidence interval would then be 756 ± 9.69, which translates to (746.31, 765.69) grams.