please please heeeeeelp 1 The UNO human resources department

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1. The UNO human resources department wanted to know about the average student employee wage rate of Omaha area. They randomly selected 30 student employees and recorded their wage rates. The sample mean and standard deviation of the wage rates are obtained as $10.2 and $3.1 respectively. Using 68-95-99.7 rule of normal distribution please answer the following questions.

a) Compute the 95% Confidence Interval (CI) for true mean wage rate (\'mu\' greek letter).

b) Interpret the confidence interval you just computed.

c) Compute the 99.7% confidence interval for true mean wage rate (mu letter)

d) For which confidence level, do you think the range (width) of interval is larger?

I) 95% CI II) 99.7% CI

e) Suppose, we construct a 75% confidence interval of true mean wage rate (mu). What do you think about the width of 75% confidence Interval (CI) compared to the width of the 95% confidence interval? I) Larger than 95% II) Smaller than 95%

f) Now if they would have selected 25 students and obtained the same mean and standard deviation of wage rate, what would be the 95% Confidence Interval (CI) for true mean wage rate (mu).

g) Is the confidence interval in a) larger than what we have in f)? Why or why not?

h) Suppose new sample standard deviation is $2.5, when 25 students were selected and mean remains the same. . What would be the 95% CI?

i) Is the confidence interval in f) larger than what we have in h)? Why or why not?

Solution

1.

a)

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 =    10.2          
z(alpha/2) = critical z for the confidence interval =    2   (empirical rule)      
s = sample standard deviation =    3.1          
n = sample size =    30          
              
Thus,              
Margin of Error E =    1.131959952          
Lower bound =    9.068040048          
Upper bound =    11.33195995          
              
Thus, the confidence interval is              
              
(   9.068040048   ,   11.33195995   ) [ANSWER]

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b)

We are 95% confident that the true mean wage rate is between $9.068 and $11.331.

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c)

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 =    10.2          
z(alpha/2) = critical z for the confidence interval =    3   (empirical rule)      
s = sample standard deviation =    3.1          
n = sample size =    30          
              
Thus,              
Margin of Error E =    1.697939928          
Lower bound =    8.502060072          
Upper bound =    11.89793993          
              
Thus, the confidence interval is              
              
(   8.502060072   ,   11.89793993   ) [ANSWER]

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d)

It is the 99.7% confidence interval that is wider, as it has greater margin of error.

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