Question

1. A researcher wishes to estimate sales per store of the Nikon D5 camera over the past year at a large chain of photography stores throughout the United States. A random and representative sample of stores that sell this camera is to be selected to provide the required estimates. To provide estimates of the population mean sales per store of this model of camera, what minimum number of stores will be necessary to sample under the following conditions?

1. The estimate desired will need to be computed with 95% confidence to within ±25 units when it is felt that the population standard deviation in the number of cameras sold per store is 50 cameras.
2. The estimate desired will now need to be computed with 90% confidence to within ±25 units when it is felt that the population standard deviation in the number of cameras sold per store is 50 cameras.
3. The estimate desired will now need to be computed with 95% confidence to within ±25 units when it is felt that the population standard deviation in the number of cameras sold per store is 65 cameras.
4. The estimate desired will now need to be computed with 95% confidence to within ±35 units when it is felt that the population standard deviation in the number of cameras sold per store is 50 cameras.

In your discussion, be sure to comment on the differences found in the calculation of the minimum sample sizes in the various parts of the above problem. Compare the results of parts b), c) and d) to the result of part a). Explain why there are differences between the minimum sample sizes in the stated parts of the problem. Give reasons for your answers that are other than purely mathematical.

Answer 1

a)

Standard Deviation ,   σ =    50                  
sampling error ,    E =   25                  
Confidence Level ,   CL=   95%                  
                          
alpha =   1-CL =   5%                  
Z value =    Zα/2 =    1.960   [excel formula =normsinv(α/2)]              
                          
Sample Size,n = (Z * σ / E )² = (   1.960   *   50   /   25   ) ² =   15.366
                          
                          
So,Sample Size needed=       16                  

b)

Standard Deviation ,   σ =    50                  
sampling error ,    E =   25                  
Confidence Level ,   CL=   90%                  
                          
alpha =   1-CL =   10%                  
Z value =    Zα/2 =    1.645   [excel formula =normsinv(α/2)]              
                          
Sample Size,n = (Z * σ / E )² = (   1.645   *   50   /   25   ) ² =   10.822
                          
                          
So,Sample Size needed=       11                  

c)

Standard Deviation ,   σ =    65                  
sampling error ,    E =   25                  
Confidence Level ,   CL=   95%                  
                          
alpha =   1-CL =   5%                  
Z value =    Zα/2 =    1.960   [excel formula =normsinv(α/2)]              
                          
Sample Size,n = (Z * σ / E )² = (   1.960   *   65   /   25   ) ² =   25.968
                          
                          
So,Sample Size needed=       26                  

d)

Standard Deviation ,   σ =    50                  
sampling error ,    E =   35                  
Confidence Level ,   CL=   95%                  
                          
alpha =   1-CL =   5%                  
Z value =    Zα/2 =    1.960   [excel formula =normsinv(α/2)]              
                          
Sample Size,n = (Z * σ / E )² = (   1.960   *   50   /   35   ) ² =   7.840
                          
                          
So,Sample Size needed=       8                  

differences between the minimum sample sizes in the stated parts of the problem is due to

1)Confidence interval: High CI means higher sample size

2)Standard deviation: High Standard deviation means higher sample size

3)Margin of Error:High Margin of Error lesser sample size

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Thanks in advance!

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Please upvote