SOLUTION: Karen wants to advertise how many chocolate chips are in each Big Chip cookie at her bakery. She randomly selects a sample of 72 cookies and finds that the number of chocolate chip

Question 1198528: Karen wants to advertise how many chocolate chips are in each Big Chip cookie at her bakery. She randomly selects a sample of 72 cookies and finds that the number of chocolate chips per cookie in the sample has a mean of 11.4 and a standard deviation of 2.7. What is the 80% 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).
Answer by proyaop(69) (Show Source): You can put this solution on YOUR website!
**1. Find the Critical Value (z-score)**
* For an 80% confidence interval, the area in each tail is (1 - 0.80) / 2 = 0.10.
* Using a standard normal distribution table or a calculator, find the z-score that corresponds to an area of 0.10 in the left tail.
* z-score ≈ 1.28
**2. Calculate the Standard Error**
* Standard Error (SE) = s / √n
* where:
* s = sample standard deviation (2.7)
* n = sample size (72)
* SE = 2.7 / √72
* SE ≈ 0.319
**3. Calculate the Margin of Error**
* Margin of Error (ME) = z* * SE
* ME = 1.28 * 0.319
* ME ≈ 0.408
**4. Calculate the Confidence Interval**
* Lower Limit = sample mean - ME = 11.4 - 0.408 = 10.992 ≈ 11.0
* Upper Limit = sample mean + ME = 11.4 + 0.408 = 11.808 ≈ 11.8
**Therefore, the 80% confidence interval for the number of chocolate chips per cookie for Big Chip cookies is (11.0, 11.8).**

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