Respuesta :
Answer:
(a) The proportion of tenth graders reading at or below the eighth grade level is 0.1673.
(b) The 95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.198, 0.260).
Step-by-step explanation:
Let X = number of students who read above the eighth grade level.
(a)
A sample of n = 269 students are selected. Of these 269 students, X = 224 students who can read above the eighth grade level.
Compute the proportion of students who can read above the eighth grade level as follows:
[tex]\hat p=\frac{X}{n}=\frac{224}{269}=0.8327[/tex]
The proportion of students who can read above the eighth grade level is 0.8327.
Compute the proportion of tenth graders reading at or below the eighth grade level as follows:
[tex]1-\hat p=1-0.8327[/tex]
[tex]=0.1673[/tex]
Thus, the proportion of tenth graders reading at or below the eighth grade level is 0.1673.
(b)
the information provided is:
n = 709
X = 546
Compute the sample proportion of tenth graders reading at or below the eighth grade level as follows:
[tex]\hat q=1-\hat p[/tex]
[tex]=1-\frac{X}{n}[/tex]
[tex]=1-\frac{546}{709}[/tex]
[tex]=0.2299\\\approx 0.229[/tex]
The critical value of z for 95% confidence interval is:
[tex]z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96[/tex]
Compute the 95% confidence interval for the population proportion as follows:
[tex]CI=\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
[tex]=0.229\pm 1.96\times \sqrt{\frac{0.229(1-0.229)}{709}}\\=0.229\pm 0.03136\\=(0.19764, 0.26036)\\\approx (0.198, 0.260)[/tex]
Thus, the 95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.198, 0.260).
