Questions: The state education commission wants to estimate the fraction of tenth grade students that have reading skills at or below the eighth grade level. Suppose a sample of 1291 tenth graders is drawn. Of the students sampled, 1098 read above the eighth grade level. Using the data, construct the 95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level. Round your answers to three decimal places.

Solution Steps

The sample proportion of tenth grade students reading at or below the eighth grade level is calculated as follows:

\[ \hat{p} = \frac{x}{n} = \frac{1291 - 1098}{1291} = \frac{193}{1291} \approx 0.149 \]

where \( x \) is the number of students reading at or below the eighth grade level, and \( n \) is the total number of students sampled.

To construct the \( 95\% \) confidence interval for the population proportion, we use the formula:

\[ \hat{p} \pm z \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \]

Here, \( z \) is the z-score corresponding to the \( 95\% \) confidence level, which is approximately \( 1.96 \).

Substituting the values:

\[ \hat{p} \pm 1.96 \cdot \sqrt{\frac{0.149(1 - 0.149)}{1291}} = 0.149 \pm 1.96 \cdot \sqrt{\frac{0.149 \cdot 0.851}{1291}} \]

Calculating the margin of error:

\[ \text{Margin of Error} = 1.96 \cdot \sqrt{\frac{0.149 \cdot 0.851}{1291}} \approx 0.019 \]

Thus, the confidence interval is:

\[ (0.149 - 0.019, 0.149 + 0.019) = (0.130, 0.169) \]

Final Answer