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.:
[tex]\hat{p}=\dfrac{235}{1123}=0.209260908281\approx0.209[/tex]
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 : [tex]\alpha= 1-0.90=0.10[/tex]
Critical value : [tex]z_{\alpha/2}=1.675[/tex]
Confidence level for population proportion:-
[tex]\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)[/tex]
Hence, the 90% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level = (0.189,0.229)
