Questions: In a sample of 10 randomly selected women, it was found that their mean height was 65.4 in. From previous studies, it is assumed that the population standard deviation for the heights of women is 2.5 in. and that the population of heights is normally distributed. Construct the 95% confidence interval for the population mean height.

Solution Steps

We have a sample of 10 randomly selected women with the following statistics:

• Sample mean height (\(\bar{x}\)): 65.4 in
• Population standard deviation (\(\sigma\)): 2.5 in
• Sample size (\(n\)): 10

We are constructing a 95% confidence interval for the population mean height. The significance level (\(\alpha\)) is calculated as: \[ \alpha = 1 - 0.95 = 0.05 \]

For a 95% confidence level, the Z-score corresponding to the critical value is approximately: \[ z \approx 1.96 \]

The standard error (SE) of the mean is calculated using the formula: \[ SE = \frac{\sigma}{\sqrt{n}} = \frac{2.5}{\sqrt{10}} \approx 0.7906 \]

The confidence interval is calculated using the formula: \[ \bar{x} \pm z \cdot SE \] Substituting the values: \[ 65.4 \pm 1.96 \cdot 0.7906 \] Calculating the margin of error: \[ 1.96 \cdot 0.7906 \approx 1.549 \] Thus, the confidence interval is: \[ (65.4 - 1.549, 65.4 + 1.549) = (63.851, 66.949) \]

Rounding the results to two decimal places, we have: \[ (63.85, 66.95) \]

Final Answer