A Management Association wishes to have information on the mean income of store

Question

A Management Association wishes to have information on the mean income of store managers in the total network of the retail industry. A random sample of 25 managers reveals a sample mean of $45,200. The Standard deviation of this population is $2,050.         a) What is the point estimate of the population mean?    b) What is the reasonable range of value for the population mean? [That is- Find the confidence interval range?]     c) How do you interpret the confidence interval range by taking for example 95 percent confidence level?

Explanation / Answer

a)

It is the sample mean,

X = 45200 [ANSWER]

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b)

Note that              
Margin of Error E = z(alpha/2) * s / sqrt(n)              
Lower Bound = X - z(alpha/2) * s / sqrt(n)              
Upper Bound = X + z(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.025          
X = sample mean =    45200          
z(alpha/2) = critical z for the confidence interval =    1.959963985          
s = sample standard deviation =    2050          
n = sample size =    25          
              
Thus,              
Margin of Error E =    803.5852337          
Lower bound =    44396.41477          
Upper bound =    46003.58523          
              
Thus, the confidence interval is              
              
(   44396.41477   ,   46003.58523   ) [ANSWER]

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c)

Hence, we are 95% confident that the true mean income of store managers in the total network of the retail industry is between $44396.41477 and $46003.58523. [CONCLUSION]

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Hi! In part b, in case you use t distirbution instead of z, here is the alternative solution:

Note that              
Margin of Error E = t(alpha/2) * s / sqrt(n)              
Lower Bound = X - t(alpha/2) * s / sqrt(n)              
Upper Bound = X + t(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.025          
X = sample mean =    45200          
t(alpha/2) = critical t for the confidence interval =    2.063898562          
s = sample standard deviation =    2050          
n = sample size =    25          
df = n - 1 =    24          
Thus,              
Margin of Error E =    846.1984103          
Lower bound =    44353.80159          
Upper bound =    46046.19841          
              
Thus, the confidence interval is              
              
(   44353.80159   ,   46046.19841   ) [ALTERNATE ANSWER, B]

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