To solve this problem, we’ll calculate a 99% confidence interval for the mean number of chocolate chips per Big Chip cookie.

Given Information:

• Sample size $n = 63$
• Sample mean $\bar{x} = 15.5$
• Sample standard deviation $s = 3.5$
• Confidence level = 99%

Step 1: Determine the Z-score

For a 99% confidence level, we look up the Z-score that corresponds to the 0.5% (0.005) in each tail, which is approximately $Z = 2.576$.

Step 2: Calculate the Standard Error

$SE = \frac{s}{\sqrt{n}} = \frac{3.5}{\sqrt{63}} \approx 0.441$

Step 3: Calculate the Margin of Error

$\text{Margin of Error} = Z \times SE = 2.576 \times 0.441 \approx 1.136$

Step 4: Calculate the Confidence Interval

$\text{Lower Limit} = \bar{x} - \text{Margin of Error} = 15.5 - 1.136 = 14.364$ $\text{Upper Limit} = \bar{x} + \text{Margin of Error} = 15.5 + 1.136 = 16.636$

Answer

The 99% confidence interval for the number of chocolate chips per cookie is: $14.364 < \mu < 16.636$

Would you like a deeper explanation on any part of this?

1. How is the Z-score determined for different confidence levels?
2. What does the standard error represent in confidence intervals?
3. Why do we use a Z-score rather than a t-score in this case?
4. How would the interval change if the confidence level were 95% instead of 99%?
5. How does sample size affect the confidence interval width?

Tip: Increasing the confidence level will make the interval wider, capturing a broader range of possible values for the mean.