Answer:
The interval is [tex]0.7187 < p < 2.421[/tex]
Step-by-step explanation:
From the question we are told that
The sample size is [tex]n = 100[/tex]
The population proportion is [tex]p = 0.81[/tex]
The confidence level is C = 98%
The level of significance is mathematically evaluated as
[tex]\alpha = 100 -98[/tex]
[tex]\alpha = 2%[/tex]%
[tex]\alpha = 0.02[/tex]
Here this level of significance represented the left and the right tail
The degree of freedom is evaluated as
[tex]df = n-1[/tex]
substituting value
[tex]df = 100 - 1[/tex]
[tex]df = 99[/tex]
Since we require the critical value of one tail in order to evaluate the 98% confidence interval that estimates the proportion of them that are involved in an after school activity. we will divide the level of significance by 2
The critical value of [tex]\frac{\alpha}{2}[/tex] and the evaluated degree of freedom is
[tex]t_{df , \alpha } = t_{99 , \frac{0.02}{2} } = 2.33[/tex]
this is obtained from the critical value table
The standard error is mathematically evaluated as
[tex]SE = \sqrt{\frac{p(1-p )}{n} }[/tex]
substituting value
[tex]SE = \sqrt{\frac{0.81(1-0.81 )}{100} }[/tex]
[tex]SE = 0.0392[/tex]
The 98% confidence interval is evaluated as
[tex]p - t_{df , \frac{\alpha }{2} } * SE < p < p + t_{df , \frac{\alpha }{2} }[/tex]
substituting value
[tex]0.81 - 2.33 * 0.0392 < p < 0.81 +2.33 * 0.0392[/tex]
[tex]0.7187 < p < 2.421[/tex]
