Suppose a sample of 2329 tenth graders is drawn. Of the students sampled, 489 read at or below 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. .Using the data, construct the 98% 98% confidence interval for the population proportion of tenth graders reading at or below the eighth grade levelupper end point:lower endpoint:

Answer:[tex]\hat p=\frac{489}{2329}=0.210[/tex]And the 98% confidence interval is (0.190;0.230). Upper end point: 0.230Lower endpoint: 0.190Step-by-step explanation:1) Data given and notationn=2329 represent the random sample taken    X represent the students who read at or below the eighth grade level[tex]\hat p=\frac{489}{2329}=0.210[/tex] estimated proportion of Americans who thinks that the Civil War is still relevant to American politics and political life in the sample  [tex]\alpha=1-0.98=0.02[/tex] represent the significance level (no given, but is assumed)    [tex]z_{\alpha/2}[/tex] represent the quantile for the normal distribution   p= population proportion of students who read at or below the eighth grade level2) SolutionThe confidence interval would be given by this formula [tex]\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex] For the 98% confidence interval the value of [tex]\alpha=1-0.98=0.02[/tex] and [tex]\alpha/2=0.01[/tex], with that value we can find the quantile required for the interval in the normal standard distribution. [tex]z_{\alpha/2}=2.33[/tex] And replacing into the confidence interval formula we got: [tex]0.210 - 2.33 \sqrt{\frac{0.21(1-0.21)}{2329}}=0.190[/tex] [tex]0.210 + 2.33 \sqrt{\frac{0.21(1-0.21)}{2329}}=0.230[/tex] And the 98% confidence interval would be given (0.190;0.230). We are confident that about 19% to 23% are tenth graders reading at or below the eighth grade level.