To calculate the 95% confidence interval for the number of chocolate chips per cookie, we can use the formula for the confidence interval for the mean when the population standard deviation is unknown and the sample size is less than 30. However, since we have a sample size of 54 (which is greater than 30), we can use the z-distribution.
The formula for the confidence interval is:
\[ \text{Confidence Interval} = \bar{x} \pm z \left( \frac{s}{\sqrt{n}} \right) \]
Where:
- \(\bar{x}\) is the sample mean,
- \(s\) is the sample standard deviation,
- \(n\) is the sample size, and
- \(z\) is the z-score corresponding to the desired confidence level (for 95%, \(z \approx 1.96\)).
Given:
- \(\bar{x} = 8.9\)
- \(s = 3.5\)
- \(n = 54\)
First, we calculate the standard error (SE):
\[ SE = \frac{s}{\sqrt{n}} = \frac{3.5}{\sqrt{54}} \approx \frac{3.5}{7.348} \approx 0.476 \]
Now, we can calculate the margin of error (ME):
\[ ME = z \cdot SE \approx 1.96 \cdot 0.476 \approx 0.933 \]
Finally, we can calculate the confidence interval:
\[ \text{Lower limit} = \bar{x} - ME = 8.9 - 0.933 \approx 7.967 \approx 8.0 \] \[ \text{Upper limit} = \bar{x} + ME = 8.9 + 0.933 \approx 9.833 \approx 9.8 \]
Thus, the 95% confidence interval for the number of chocolate chips per cookie is:
\[ 8.0 < \mu < 9.8 \]
So the answer is:
8.0 < mu < 9.8
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