Karen wants to advertise how many chocolate chips are in each Big Chip cookie at her bakery. She randomly selects a sample of 43 cookies and finds that the number of chocolate chips per cookie in the sample has a mean of 18.9 and a standard deviation of 3.8. What is the $98 \%$ confidence interval for the number of chocolate chips per cookie for Big Chip cookies? Enter your answers accurate to one decimal place (because the sample statistics are reported accurate to one decimal place). \[ 1.3 \quad x<\mu<20.2 \quad x \] Question Help: $\square$ Video $\square$ Message instructor

Step 1 ：The problem is asking for a 98% confidence interval for the number of chocolate chips per cookie. The confidence interval can be calculated using the formula for a confidence interval which is: \(\bar{x} \pm Z \frac{s}{\sqrt{n}}\)

Step 2 ：In this formula, \(\bar{x}\) is the sample mean, \(Z\) is the Z-score which corresponds to the desired level of confidence, \(s\) is the sample standard deviation, and \(n\) is the sample size.

Step 3 ：For this problem, \(\bar{x} = 18.9\), \(s = 3.8\), and \(n = 43\). The Z-score for a 98% confidence interval is approximately 2.33.

Step 4 ：Substitute these values into the formula to calculate the confidence interval: \(18.9 \pm 2.33 \frac{3.8}{\sqrt{43}}\)

Step 5 ：The margin of error is approximately 1.35.

Step 6 ：The lower bound of the confidence interval is \(18.9 - 1.35 = 17.55\) and the upper bound is \(18.9 + 1.35 = 20.25\).

Step 7 ：Rounding to one decimal place, the 98% confidence interval for the number of chocolate chips per cookie for Big Chip cookies is \(\boxed{[17.5, 20.3]}\).