Question

Part 1: Analyzing your College’s School Graduation Rate

You recently went through your college website and some information there got your attention. There was a claim that your college has a 77% graduation rate. You thought it would be interesting to check the validity of this statement since these days you are reading about hypothesis testing in your Statistics online course. You contacted the research department and got access to the data for the last graduation and out of 200 students 165 graduated.

To complete Case 1 please answer the following questions:

1. If you are to conduct a hypothesis test using the data above,
   1. Which test (t-test or z test) would you think be appropriate and why?

2. Is this should be a test on population proportion or mean?

3. Calculate the sample statistics (sample mean or sample proportion based on the information given) (1 mark)

2. Conduct the hypothesis test with a 0.01 level of significance. You may use ‘the percentage of graduates not equals to 0.77’ as your alternative hypothesis. Please show your work, the 5 steps as described in the textbook.

3. Construct a 95% confidence interval for the population proportion of graduates. Show your work.

Answer 1

(a)

(I) z- test because to test for proportions, z test is used

(II) Test on population proportion

(III) n = 200, x = 165, p' = x/n = 0.825

(b)

Data:    

n = 200   

p = 0.77   

p' = 0.825   

Hypotheses:    

Ho: p = 0.77   

Ha: p ≠ 0.77   

Decision Rule:    

α = 0.01   

Lower Critical z- score = -2.5758   

Upper Critical z- score = 2.5758   

Reject Ho if |z| > 2.5758   

Test Statistic:    

SE = √{p (1 - p)/n} = √(0.77 * (1 - 0.77)/200) = 0.0298

z = (p'- p)/SE = (0.825 - 0.77)/0.029757352032733 = 1.8483

p- value = 0.0646   

Decision (in terms of the hypotheses):    

Since 1.8483 < 2.5758 we fail to reject Ho

Conclusion (in terms of the problem):    

There is no sufficient evidence that the population percentage is not 77%.   

(c)

n = 200    

p = 0.825    

% = 95    

Standard Error, SE = √{p(1 - p)/n} =    √(0.825(1 - 0.825))/200 = 0.026867732

z- score = 1.959963985    

Width of the confidence interval = z * SE =     1.95996398454005 * 0.0268677315752558 = 0.05265979

Lower Limit of the confidence interval = P - width =     0.825 - 0.0526597862337911 = 0.77234021

Upper Limit of the confidence interval = P + width =     0.825 + 0.0526597862337911 = 0.87765979

The 95% confidence interval is [0.7723, 0.8777]