Question

Past experience indicates that because of low morale, a company loses 20 hours a year per employee due to lateness and abstenteeism. Assume that the standard deviation of the population is 6 and normally distributed.

The HR department implemented a new rewards system to increase employee morale, and after a few months it collected a random sample of 20 employees and the annualized absenteeism was 14.

1. Could you confirm that the new rewards system was effective with a 90% confidence?
2. An HR subject matter expert would be very happy if the program could reduce absenteeism by 20% (i.e. to 16 hours). Given the current sampling parameters, what is the probability that the new rewards system reduced absenteeism to 16 hours and you miss it?
3. Repeat part 1) and 2) with an α = 95% CI.
4. Based on the answers in 3), is the sampling method good enough to identify a reduction from 20 to 16 hours if I use a confidence of 95%?
5. What should the sample size be if you want β to be 5%

Answer 1

1)

Level of Significance ,    α =    0.1          
'   '   '          
z value=   z α/2=   1.6449   [Excel formula =NORMSINV(α/2) ]      
                  
Standard Error , SE = σ/√n =   6.0000   / √   20   =   1.341641
margin of error, E=Z*SE =   1.6449   *   1.34164   =   2.206803
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    14.00   -   2.206803   =   11.793197
Interval Upper Limit = x̅ + E =    14.00   -   2.206803   =   16.206803
90%   confidence interval is (   11.79   < µ <   16.21   )

As you can 20 do not fall in above interval and we can say that the new rewards system was effective with a 90% confidence

2)

Zα/2   = ±   1.960   (two tailed test)                      
We will fail to reject the null (commit a Type II error) if we get a Z statistic between                           -1.960   and   1.960
these Z-critical value corresponds to some X critical values ( X critical), such that                                  
                                  
-1.960   ≤(x̄ - µo)/σx≤   1.960                          
17.370   ≤ x̄ ≤   22.630                          
                                  
now, type II error is ,ß =        P (   17.370   ≤ x̄ ≤   22.630   )          
       Z =    (x̄-true mean)/σx                      
       Z1 = (   17.370   -   16   ) /   1.34164   =   1.021
       Z2 = (   22.630   -   16   ) /   1.34164   =   4.941
                                  
   so, P(   1.021   ≤ Z ≤   4.941   ) = P ( Z ≤   4.941   ) - P ( Z ≤   1.021   )
                                  
       =   1.000   -   0.846   =   0.1535 [ Excel function: =NORMSDIST(z) ]  

3)

Yes we can use 95% interval

Sample size should atleat be 30

Thanks in advance!

revert back for doubt

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