Answer :
Answer:
The estimation for the proportion of tenth graders reading at or below the eighth grade level is given by:
[tex] \hat p =\frac{955-812}{955}= 0.150[/tex]
[tex]0.150 - 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.131[/tex]
[tex]0.150 + 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.169[/tex]
And the 90% confidence interval would be given (0.131;0.169).
Step-by-step explanation:
We have the following info given:
[tex]n= 955[/tex] represent the sampel size slected
[tex] x = 812[/tex] number of students who read above the eighth grade level
The estimation for the proportion of tenth graders reading at or below the eighth grade level is given by:
[tex] \hat p =\frac{955-812}{955}= 0.150[/tex]
The confidence interval for the proportion would be given by this formula
[tex]\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
For the 90% confidence interval the significance is [tex]\alpha=1-0.9=0.1[/tex] and [tex]\alpha/2=0.05[/tex], with that value we can find the quantile required for the interval in the normal standard distribution and we got.
[tex]z_{\alpha/2}=1.64[/tex]
And replacing into the confidence interval formula we got:
[tex]0.150 - 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.131[/tex]
[tex]0.150 + 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.169[/tex]
And the 90% confidence interval would be given (0.131;0.169).