Karen wants to advertise how many chocolate chips are in each Big Chip cookie at her bakery. She randomly selects a sample of 64 cookies and finds that the number of chocolate chips per cookie in the sample has a mean of 6.1 and a standard deviation of 3.4 . Construct a \( 90 \% \) confidence interval to estimate the average number of chocolate chips per cookie for Big Chip cookies? Enter your answers accurate to one decimal place.

To find the \(90\%\) confidence interval for the average number of chocolate chips per cookie, we first need the standard error of the mean (SEM), which is calculated using the standard deviation (SD) and the sample size (n). 1. Calculate the standard error (SEM): \[ SEM = \frac{SD}{\sqrt{n}} = \frac{3.4}{\sqrt{64}} = \frac{3.4}{8} = 0.425 \] 2. Determine the critical value for a \(90\%\) confidence level. For a two-tailed test with \(n-1 = 63\) degrees of freedom, the critical value (from Z-tables or using a calculator) is approximately \(1.645\). 3. Now calculate the margin of error (ME): \[ ME = Z \times SEM = 1.645 \times 0.425 \approx 0.6981 \] 4. Finally, construct the confidence interval: \[ CI = \text{mean} \pm ME = 6.1 \pm 0.6981 \] So, \[ CI = (6.1 - 0.6981, 6.1 + 0.6981) = (5.4, 6.8) \] Rounded to one decimal place, the \(90\%\) confidence interval for the average number of chocolate chips per cookie is approximately \( (5.4, 6.8) \).