The university hires you to determine whether the current students at UWC receiv

Question

The university hires you to determine whether the current students at UWC receive more parking tickets than the historical average.

In years past, the average UWC student earns $115 in parking tickets over a 4-year period. You take a random sample of 113 out-going seniorsto determine if they earn more or less in parking tickets.

Your sample has a mean of $124 and a standard deviation of $45.

A.) Given our p-value can we reject the null hypothesis that there is no difference in the amount owed in parking tickets between our sample of UWC students and the historical average of $115?

True

False

B.) So...does our sample of 113 UWC seniors come from a population with an average parking ticket expense of $115?

True

False

C.) We can be 95% confident that the population mean for our UWC sample is between ____ and ____?

Explanation / Answer

a)

Formulating the null and alternative hypotheses,              
              
Ho:   u   <=   115  
Ha:    u   >   115  
              
As we can see, this is a    right   tailed test.      
              
              
Getting the test statistic, as              
              
X = sample mean =    124          
uo = hypothesized mean =    115          
n = sample size =    113          
s = standard deviation =    45          
              
Thus, z = (X - uo) * sqrt(n) / s =    2.126029163          
              
Also, the p value is              
              
p =    0.016750416          
              
As P < 0.05, we   REJECT THE NULL HYPOTHESIS. [TRUE]

*************************

B)

FALSE, from the conclusion in part A.

***************************  
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 =    124          
z(alpha/2) = critical z for the confidence interval =    1.959963985          
s = sample standard deviation =    45          
n = sample size =    113          
              
Thus,              
Margin of Error E =    8.297005597          
Lower bound =    115.7029944          
Upper bound =    132.2970056          
              
Thus, the confidence interval is              
              
(   115.7029944   ,   132.2970056   ) [ANSWER]

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