Question

Using the dataset NutritionStudy, we calculate that the average number of grams of fat consumed in a day for the sample of \(n=315\) US adults in the study is \(\bar{x}=77.03\) grams with \(s=33.83\) grams. (a) Find and interpret a \(95 \%\) confidence interval for the average number of fat grams consumed per day by US adults. (b) What is the margin of error? (c) If we want a margin of error of only ±1 , what sample size is needed?

Short answer

The 95% confidence interval is (75.00, 79.06) grams. The margin of error is 2.03 grams. A sample size of approximately 4435 is required to achieve a margin of error of ±1.

Step-by-step solution

Calculate the Confidence Interval

Make use of the given sample mean, standard deviation, and sample size to calculate the 95% confidence interval. The formula for a confidence interval is as follows: \[\bar{x} \pm z \times \frac{s}{\sqrt{n}}\], Where: \(\bar{x}\) is the sample mean, \(z\) is the z-score, \(s\) is the standard deviation, and \(n\) is the sample size. For a 95% confidence interval, \(z\) is approximately 1.96.

Compute Margin of Error

Calculate the margin of error. The margin of error is basically the product of the z-score and the standard error, and it provides an estimate of how much the sampled mean is expected to vary from the real population mean. The formula is: \(E = z \times \frac{s}{\sqrt{n}}\).

Determining the Required Sample Size

To obtain a margin of error of only ±1, calculate the required sample size. The formula to obtain this is: \(n = \left(\frac{z \times s}{E}\right)^2\), with \(E\) being the desired margin of error.