Question

The state education commission wants to estimate the fraction of tenth grade students that have reading skills at or below the eighth grade level.a.) Suppose a sample of 1123 tenth graders is drawn. Of the students sampled, 888 read above the eighth grade level. Using the data, estimate the proportion of tenth graders reading at or below the eighth grade level. Enter your answer as a fraction or a decimal number rounded to three decimal places.b.) Suppose a sample of 1123 tenth graders is drawn. Of the students sampled, 888 read above the eighth grade level. Using the data, construct the 90% 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 1

Answer with explanation:

a) Given : Sample size of students : n =1123

Number of students read above the eighth grade level = 888

Then, Number of students read at or below the eighth grade level =1123-888=235

Then , the sample proportion of tenth graders reading at or below the eighth grade level.:

$\hat{p}=\dfrac{235}{1123}=0.209260908281\approx0.209$

Since, the best estimate for the population proportion is the sample proportion.

Thus, the estimated proportion of tenth graders reading at or below the eighth grade level = 0.209

b) Significance level : $\alpha= 1-0.90=0.10$

Critical value : $z_{\alpha/2}=1.675$

Confidence level for population proportion:-

$\hat{p}\pm z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}\\\\=0.209\pm (1.645)\sqrt{\dfrac{0.209(1-0.209)}{1123}}\\\\=0.209\pm0.0199589370751\\\\\approx 0.209\pm0.020=(0.189,0.229)$

Hence, the 90% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level = (0.189,0.229)