To calculate a 90% confidence interval for the population mean, we can use the formula for the confidence interval when the population standard deviation is known:
\[ CI = \bar{x} \pm z \left(\frac{\sigma}{\sqrt{n}}\right) \]
where:
- \(\bar{x}\) is the sample mean
- \(z\) is the z-score corresponding to the desired confidence level
- \(\sigma\) is the population standard deviation
- \(n\) is the sample size
Given:
- Sample mean, \(\bar{x} = 68\)
- Population standard deviation, \(\sigma = 3\)
- Sample size, \(n = 25\)
First, find the z-score for a 90% confidence level. The critical z-value that corresponds to 90% confidence level (which has 5% in each tail of the standard normal distribution) is approximately 1.645.
Now, we compute the standard error (SE):
\[ SE = \frac{\sigma}{\sqrt{n}} = \frac{3}{\sqrt{25}} = \frac{3}{5} = 0.6 \]
Now we can calculate the margin of error (ME):
\[ ME = z \cdot SE = 1.645 \cdot 0.6 \approx 0.987 \]
Now we can find the confidence interval:
\[ CI = \bar{x} \pm ME = 68 \pm 0.987 \]
Calculating the lower and upper bounds:
\[ \text{Lower bound} = 68 - 0.987 \approx 67.013 \] \[ \text{Upper bound} = 68 + 0.987 \approx 68.987 \]
Finally, rounding to the nearest hundredths, we get:
\[ CI \approx (67.01, 68.99) \]
Thus, the 90% confidence interval estimate for the population mean of exam scores is:
\[ (67.01, 68.99) \]
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