A nurse in a large university N30000 is concerned about stud

A nurse in a large university (N=30000) is concerned about students eye health. She takes a random sample of 75 students who don’t wear glasses and finds 27 that need glasses.

What the point estimate of p, the population proportion?

Whats the critical z value for a 90% confidence interval for the population proportion?

Whats the margin of error for a 90% confidence interval for the population proportion?

Calculate the 90% confidence interval for the population proportion.

Using your graphing calculator find a 95% confidence interval for the proportion of students who need to wear glasses but done. Show all work.

The nurse wants to be able to estimate, with a 95% confidence interval and a margin of error of 6% the proportion of students who need to wear glasses but don’t. Fine the necessary sample size (n) for this estimate.

Solution

a)

p^ = 27/75 = 0.36 [answer]

b)

Using a table or technology,

zcrit = 1.644853627 [answer]

c)

E = zcrit * sqrt(p^ (1 - p^) / n) = 1.644853627*sqrt(0.36*(1-0.36)/75) = 0.091167042 [answer]

d)

The confidence interval is thus,

(p^-E, p^+E) = (0.268832958, 0.451167042)

e)

Note that      
      
n = z(alpha/2)^2 p (1 - p) / E^2      
      
where      
      
alpha/2 =    0.025  
       
      
Using a table/technology,      
      
z(alpha/2) =    1.959963985  
      
Also,      
      
E =    0.06  
p =    0.36  
      
Thus,      
      
n =    245.8533645  
      
Rounding up,      
      
n =    246   [ANSWER]