Question

suppose a sample of 1305 tenth graders is drawn. Of the students sampled, 1044 read above the eighth grade level. Using the data, construct the 99% 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: $(0.771,\ 0.829)$

Step-by-step explanation:

Given : Significance level : $\alpha: 1-0.99=0.01$

Critical value : $z_{\alpha/2}=2.576$

Sample size : n=1305

The proportion of tenth graders read above the eighth grade level. =$p=\dfrac{1044}{1305}=0.8$

The confidence interval for population proportion is given by :-

$p\pm z_{\alpha/2}\sqrt{\dfrac{p(1-p)}{n}}\\\\=0.8\pm(2.576)\sqrt{\dfrac{0.8(1-0.8)}{1305}}\\\\\approx0.8\pm0.029\\\\=(0.771,\ 0.829)$

Hence, the 99% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.771,0.829) .