To calculate the 90% confidence interval for the population mean, we can use the formula:
\[ \bar{x} \pm z \left( \frac{\sigma}{\sqrt{n}} \right) \]
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 \(n = 100\)
- For a 90% confidence level, the z-score (from z-tables or standard normal distribution) is approximately 1.645.
Now, we can calculate:
- Calculate the standard error (SE):
\[ SE = \frac{\sigma}{\sqrt{n}} = \frac{3}{\sqrt{100}} = \frac{3}{10} = 0.3 \]
- Calculate the margin of error (ME):
\[ ME = z \cdot SE = 1.645 \cdot 0.3 \approx 0.4935 \]
- Calculate the confidence interval:
\[ \text{Lower limit} = \bar{x} - ME = 68 - 0.4935 \approx 67.5065 \] \[ \text{Upper limit} = \bar{x} + ME = 68 + 0.4935 \approx 68.4935 \]
Rounding to the nearest hundredths, we have:
\[ \text{Lower limit} \approx 67.51 \] \[ \text{Upper limit} \approx 68.49 \]
Thus, the 90% confidence interval estimate for the population mean of exam scores is:
\[ (67.51, 68.49) \]
Final answer:
(67.51,68.49)