The Professor's Response

For the question below, I got 0.001 and 0.049. The solution has 0.003, 0.047. Can you explain how to do this one?

Assuming that grades in a statistics class are normally distributed with a mean of 75 and a standard deviation of 20, there should be around 5% of the students who earn a C-. In a recent study of a class of 200 students there were 5 grades of C-. Construct the 95% confidence interval for the true proportion of C- grades for statistics classes like the class used in the study 200. If the class grades were normally distributed with a mean of 75 and a standard deviation of 20, we'd expect that around 10 grades (5% of 200) would be in the C- range. Looking at the interval we just created, what do you think about the idea that the grades are normally distributed?

  See the professor's answer below.

Professor McGukian

Hi Eric,
This is a proportion CI problem, so we need the following data items:
n = 200
p-hat = 5/200 = .025
q-hat = 1 - .025 = .975
CL = 95%

Then we get Za/2 = 1.960 (look up 0.025 in one-tail on the t-table for a large sample size)

Find the margin of error:
E = za/2*root(p-hat*q-hat/n) = 1.960*root(.025*.975/200) = 0.02164

Fill in (p-hat - E, p-hat + E) = (.025 - .02164,  .025+ .02164) = (0.003, 0.047)

Since the interval does not contain 5% as a candidate for the true proportion, perhaps the data is not normally distributed. 

Hope that helps,

Professor McGuckian

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