Question

There are 1254 machinery rebuilding and repairing companies in the United States. A tool manufacturer wishes to survey a simple random sample of these firms to find out what proportion of them are interested in a new tool design. Assume that we’re dealing with a finite population in the following sub-questions.

(a) If the tool manufacturer would like to be 95% confident that the sample proportion is within 0.01 of the actual population proportion, how many machinery rebuilding and repairing companies should be included in the sample (Hint: use the conservative value of p for when p is not given)?

b) Suppose the tool manufacturer has carried out the study, using the sample size determined in part (a), and 39.0% of the machinery rebuilding and repairing companies are interested in the new tool design. Which of the following 95% confidence interval for the population percentage? Show work.

A. CI = (0.3710, 0.4091)

B. CI = (0.3613, 0.4187)

C. CI = (0.3572, 0.4228)

D. CI = (0.3338, 0.4462)

(c)Which of the following is the best interpretation of the confidence interval from the previous part?

A. After performing a large number of samples, we expect to arrive at the same confidence interval as above 95% of the time.

B. We are 95% confident that the unknown population parameter, π, falls in this interval.

C. 95% of all the data values in the population fall within the interval.

D. There is a 5% margin of error in our statistical analysis.

Answer 1

a)

without prior estimate, let    sample proportion ,   p̂ =    0.5                          
   sampling error ,    E =   0.01                          
   Confidence Level ,   CL=   0.95                          
                                      
   alpha =   1-CL =   0.05                          
   Z value =    Zα/2 =    1.960   [excel formula =normsinv(α/2)]                      
                                      
   Sample Size,n =   p̂ * (1-p̂) /((E / Z)² + p̂(1-p̂)/N) = 0.50   * ( 1 -   0.50   )/((0.01/1.96)²+0.5*0.5/1254) ≈1109.17
                                      
                                      
   so,Sample Size required= 1110

b)

Level of Significance,   α =    0.05          
Sample Size,   n =    1110          
                  
Sample Proportion ,    p̂ = 0.3900          
z -value =   Zα/2 =    1.960   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.0146          
margin of error , E = Z*SE =    1.960   *   0.0146   =   0.0287
                  
95%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.390   -   0.0287   =   0.361
Interval Upper Limit = p̂ + E =   0.390   +   0.0287   =   0.419
                  
95%   confidence interval is (   0.3613   < p <    0.4187   )

c)

B. We are 95% confident that the unknown population parameter, π, falls in this interval.