Government officials in Canberra have recently expressed concern regarding overr

Question

Government officials in Canberra have recently expressed concern regarding overruns on military contracts. These unplanned expenditures have been costing Australians millions of dollars every year. The prime minister impanels a committee of experts to estimate the average amount each contract costs the government over and above the amount agreed upon. The committee has already determined that the standard deviation in overruns is $17.5 million, and that they appear normally distributed. i. If a sample of 25 contracts is selected, how likely is it the sample will overestimate the population mean by more than $10 million? ii. The prime minister will accept an error of $5 million in the estimate of mu. How likely is he to receive an estimate from the committee within the specified range?

Explanation / Answer

i.

Note that the mean deviation from the population mean is 0.

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    10      
u = mean =    0      
n = sample size =    25      
s = standard deviation =    17.5      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    2.857142857      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   2.857142857   ) =    0.002137367 [ANSWER]

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ii.

Hence, an error from -5 to 5 million.

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    -5      
x2 = upper bound =    5      
u = mean =    0      
n = sample size =    25      
s = standard deviation =    17.5      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    -1.428571429      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    1.428571429      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.076563726      
P(z < z2) =    0.923436274      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.846872549   [ANSWER]  

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