Karen wants to advertise how many chocolate chips are in each Big Chip cookie at her bakery. She randomly selects a sample of 54 cookies and finds that the number of chocolate chips per cookie in the sample has a mean of 8.9 and a standard deviation of 3.5. What is the 95% 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).

Question

GPT-4o mini · Answer

To calculate the 95% confidence interval for the number of chocolate chips per cookie, we can use the formula for the confidence interval of the mean:

$Confidence Interval = \bar{x} \pm z (\frac{s}{\sqrt{n}})$

where:

$\bar{x}$ is the sample mean,
$z$ is the z-value for the desired confidence level,
$s$ is the sample standard deviation,
$n$ is the sample size.

From the problem:

$\bar{x} = 8.9$
$s = 3.5$
$n = 54$

Step 1: Find the z-value for a 95% confidence level

For a 95% confidence level, the z-value (the critical value) from the standard normal distribution is approximately 1.96.

Step 2: Calculate the standard error (SE)

The standard error (SE) is calculated as follows:

$S E = \frac{s}{\sqrt{n}} = \frac{3.5}{\sqrt{54}} \approx \frac{3.5}{7.348} \approx 0.476$

Step 3: Calculate the margin of error (ME)

Now, calculate the margin of error:

$M E = z \cdot S E = 1.96 \cdot 0.476 \approx 0.93456$

Step 4: Calculate the confidence interval

Using the margin of error, the confidence interval is:

$Confidence Interval = \bar{x} \pm M E = 8.9 \pm 0.93456$

Calculating the lower and upper bounds:

$Lower Bound = 8.9 - 0.93456 \approx 7.96544 (approximately 8.0)$

$Upper Bound = 8.9 + 0.93456 \approx 9.83456 (approximately 9.8)$

Conclusion

Thus, the 95% confidence interval for the number of chocolate chips per cookie is:

$(8.0, 9.8)$