The amount of weight a high school student carries in his or her bookbag follows a normal distribution with a standard deviation of σ = 2.5 pounds. Suppose a random sample of 22 bookbags produced a mean of 15 pounds. Construct a 95% confidence interval to estimate the mean bookbag weight for all high school students.

Answer:
Given:

Sample Mean $ (x\bar)=15 $

Population Standard Deviation $ (\sigma)=2.5 $

Sample Size $ (n) = 22 $

Confidence Interval Level $ (CI) = 95\% $

Solution:

The level of significance $ (\alpha):$

$ \alpha = 1 – 0.95 = 0.05 $

The critical value $ (Z_c): $

$ Z_c = Z_{\alpha/2} = Z_{0.05/2} = 1.96 $

The confidence interval $ (CI): $

$\text{CI} = \bar{x} \pm Z_c \cdot \frac{\sigma}{\sqrt{n}} $

$ = 15 \pm 1.96 \cdot \frac{2.5}{\sqrt{22}} $

$ = (13.961, 16.039) $

Final Answer:

The 95% confidence interval $ = (13.961, 16.039) $