Question

A website is trying to increase registration for first-time visitors, exposing 1% of these visitors to a new site design. Of 752 randomly sampled visitors over a month who saw the new design, 64 registered.

(a) Check any conditions required for constructing a confidence interval.

(b) Compute the standard error.

(c) Construct and interpret a 90% confidence interval for the fraction of first-time visitors of the site who would register under the new design (assuming stable behaviors by new visitors over time).

Answer 1

a)

np=752*0.01=7.52 ≥5

n(1-p) = 752*(1-0.01) = 744.48 ≥5

At least 5 expected successes and 5 expected failures in sample. Normality assumption okay.              

sample size,n < 5% of population size,N
so, all conditions are satisfied.

b)

Level of Significance,   α =    0.10          
Number of Items of Interest,   x =   64          
Sample Size,   n =    752          
                  
Sample Proportion ,    p̂ = x/n =    0.085          
z -value =   Zα/2 =    1.645   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.0102          

c)

margin of error , E = Z*SE =    1.645   *   0.0102   =   0.0167
                  
90%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.085   -   0.0167   =   0.0684
Interval Upper Limit = p̂ + E =   0.085   +   0.0167   =   0.1018
                  
90%   confidence interval is (   0.0684   < p <    0.1018   )