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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

Hi! I am doing number 48 and for some reason I don't get the answer posted on the hw, i get 1.07, 5.13. However, when I do it on my graphing calculator like you did in the video, I get the right answer. I was wondering if you had a video of you doing the problem w/o the graphing calculator so I could see where I am doing it wrong.

Posted to **STATS 2** on Sunday, September 14, 2014 Replies: 2

09/14/2014

3:00 PM EST

Hi Ruby,

The mistake you are making is that you are not using the t critical value like you are supposed to use. You are using the z critical value. In step 2 of the problem, you are going to the t table and looking up 0.005 in one tail (for 99% confidence), which is correct, but then you go straight to the bottom of the table to get the critical z value 2.576.

That is your error. Instead of going to the bottom of the t table, you should be stopping in the 28 degrees of freedom row because your sample size is only 29. Remember for 30 or less, we will use the t critical value with n - 1 degrees of freedom, not the z critical value which is found in the last row of the table.

If you use the correct critical value, your margin of error will look like this:

2.763 * 4.25/SquareRoot(29) = 2.18057

Then you can add and subtract this margin of error to and from the mean to get the right interval.

Hope that makes sense,

Professor McGuckian

09/14/2014

3:53 PM EST

Thank you so much!!!!!!!!!