Question

A company issues a survey in order to find out whether its employees feel stressed at work. It randomly selects 200 employees and sends them the survey. 145 employees return a completed survey, 56 of whom say they feel stressed.

(a) Construct a confidence interval for the proportion of employees in the company who are feeling stressed.

(b) If more employees had returned a completed survey, would the confidence interval be wider or narrower than your answer in (a)? Give reasons for your answer.

(c) Comment on the sampling method used by the company in this case. How could it

be improved?

Answer 1

a)

Level of Significance,   α =    0.05          
Number of Items of Interest,   x =   56          
Sample Size,   n =    145          
                  
Sample Proportion ,    p̂ = x/n =    0.386          
z -value =   Zα/2 =    1.960   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.0404          
margin of error , E = Z*SE =    1.960   *   0.0404   =   0.0792
                  
95%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.386   -   0.0792   =   0.3070
Interval Upper Limit = p̂ + E =   0.386   +   0.0792   =   0.4655
                  
95%   confidence interval is (   0.3070   < p <    0.4655   )

b)

More sample means less margin of error.

Less margin of error means less wider the confidence interval

c)

Random Sampling has been used.

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