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Course Documents

Chapter 1 - Intro

Chapter 2 - Methods for Describing Sets of Data

Chapter 3 - Probability

Chapter 4 - Discrete Random Variables

Chapter 5 - Normal Random Variables

Chapter 6 - Sampling Distributions

Chapter 7 - Confidence Intervals

Chapter 8 - Tests of Hypothesis: One Sample

Chapter 9 - Confidence Intervals and Hypothesis Tests: Two Samples

Sample Exam I: Chapters 1 & 2

Sample Exam II: Chapters 3 & 4

Sample Exam III: Chapters 5 & 6

Sample Exam IV: Chapters 7 & 8

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?

Posted to **STATS 2** on Wednesday, August 14, 2013 Replies: 1

08/15/2013

1:09 AM EST

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