Assume that a random sample of n25 Western Michigan Universi

Assume that a random sample of n=25 Western Michigan University students resulted in a mean summer income of $3,500 with a standard deviation of $1,000. Based on this information, calculate the 95% Confidence Interval for the mean summer income of ALL Western Michigan University students.

A random sample of n=36 Western Michigan University students shows a mean student loan debt of $15,000 with a standard deviation of $6,000. Based on this information, what is the probability that the mean student loan debt of Western Michigan University students is less than $17,000?

In 2000, the 95% Confidence Interval for the mean age of a WMU student\'s car was found to be 5 to 9 years old (5, 9). In 2012, a random sample of WMU students found the mean age of a student\'s car to be 10 years old. Is there reason to believe that the mean age of a WMU student\'s car has changed from what is was in 2000?

A local TV station did some exit polling of voters for a particular ballot proposal.

A 95% Confidence Interval for the proportion of voters in favor of this ballot proposal was (.45, .55). Based on these results, what should the TV station report to its viewers regarding the status of this proposal?

Solution

1.

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 =    3500          
t(alpha/2) = critical t for the confidence interval =    2.063898562          
s = sample standard deviation =    1000          
n = sample size =    25          
df = n - 1 =    24          
Thus,              
Margin of Error E =    412.7797123          
Lower bound =    3087.220288          
Upper bound =    3912.779712          
              
Thus, the confidence interval is              
              
(   3087.220288   ,   3912.779712   ) [ANSWER]

2.

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    17000      
u = mean =    15000      
n = sample size =    36      
s = standard deviation =    6000      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    2      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   2   ) =    0.977249868 [ANSWER]

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