Save \& Exit Certify Lesson: 8.4 Estimating Population Proporti. NYAH GRIME Question 3 of 8, Step 2 of 2 $4 / 13$ Correct The state education commission wants to estimate the fraction of tenth grade students that have reading skills at or below the eighth grade level Step 2 of 2 : Suppose a sample of 929 tenth graders is drawn. Of the students sampled, 185 read at or below the eighth grade level. Using the data, construct the $98 \%$ confidence interval for the population proportion of tenth graders reading at or below the eighth grade level Round your answers to three decimal places. Answer How to enter your answer (opens in new window) Tables: Keypad Kevbo rd Shortcuts Previous step answer: Lower endpoint Upper endpoint
Solution
Step 1 ：Given that the sample size \(n = 929\) and the number of successes \(x = 185\). The Z-score for the desired confidence level of 98% is approximately \(Z = 2.33\).
Step 2 ：First, we calculate the sample proportion \(\hat{p}\) which is the number of successes divided by the sample size. \(\hat{p} = \frac{x}{n} = \frac{185}{929} = 0.199\).
Step 3 ：Next, we calculate the standard error (SE) using the formula \(\sqrt{ \frac{\hat{p}(1-\hat{p})}{n} }\). Substituting the values, we get \(SE = \sqrt{ \frac{0.199*(1-0.199)}{929} } = 0.013\).
Step 4 ：Finally, we calculate the confidence interval (CI) using the formula \(\hat{p} \pm Z * SE\). Substituting the values, we get \(CI_{lower} = 0.199 - 2.33 * 0.013 = 0.169\) and \(CI_{upper} = 0.199 + 2.33 * 0.013 = 0.230\).
Step 5 ：The 98% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is approximately \(\boxed{[0.169, 0.230]}\).
