SOLUTION: Aisha wants to advertise how many chocolate chips are in each Big Chip cookie at her bakery. She randomly selects a sample of 79 cookies and finds that the number of chocolate chip

Question 1198529: Aisha wants to advertise how many chocolate chips are in each Big Chip cookie at her bakery. She randomly selects a sample of 79 cookies and finds that the number of chocolate chips per cookie in the sample has a mean of 19.2 and a standard deviation of 2.9. What is the 90% confidence interval for the number of chocolate chips per cookie for Big Chip cookies? Assume the data is from a normally distributed population. Round answers to 3 decimal places where possible.
Answer by proyaop(69) (Show Source): You can put this solution on YOUR website!
**1. Find the Critical Value (z-score)**
* For a 90% confidence interval, the critical value (z*) is 1.645. This value corresponds to the z-score that captures 90% of the data within two standard deviations of the mean in a standard normal distribution.
**2. Calculate the Standard Error**
* Standard Error (SE) = s / √n
* where:
* s = sample standard deviation (2.9)
* n = sample size (79)
* SE = 2.9 / √79
* SE ≈ 0.326
**3. Calculate the Margin of Error**
* Margin of Error (ME) = z* * SE
* ME = 1.645 * 0.326
* ME ≈ 0.536
**4. Calculate the Confidence Interval**
* Lower Limit = sample mean - ME = 19.2 - 0.536 = 18.664
* Upper Limit = sample mean + ME = 19.2 + 0.536 = 19.736
**Therefore, the 90% confidence interval for the number of chocolate chips per cookie for Big Chip cookies is (18.664, 19.736).**
This means that Aisha can be 90% confident that the true average number of chocolate chips per cookie in all her Big Chip cookies falls within this range.

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