Teaching - Statistics

Last Day - 7 December 2006

noreply@blogger.com (Murali Shanker) — Sun, 10 Dec 2006 18:38:00 +0000

Mini Exam - 5 December 2006

noreply@blogger.com (Murali Shanker) — Thu, 07 Dec 2006 20:20:00 +0000

Here are the recordings for the mini exam we had in class.

Section 001
Section 002

Also, for Section 001, here are the last few pages for Chapter 13.

Section 001

Chapter 13: 28 November 2006

noreply@blogger.com (Murali Shanker) — Thu, 30 Nov 2006 02:44:00 +0000

We continued our discussion on Linear Regression and Correlation. We saw that the linear regression models are generally valid only for the range of data observed.

Section 001
Section 002

Chapter 13: 16 November 2006

noreply@blogger.com (Murali Shanker) — Fri, 17 Nov 2006 03:54:00 +0000

We started our discussion on Linear Regression and Correlation. We saw examples of scatter plots, and correlation coefficients.

Section 001
Section 002

Additional Review Session

noreply@blogger.com (Murali Shanker) — Wed, 15 Nov 2006 04:32:00 +0000

Some students have expressed an interest in having an additional review session. I can do that on the following days and time:

This Friday (11/17): 2 - 3
Weekend: Usually in the afternoon
Monday: 1-3

Please leave your comments if you are interested, and what times are suitable.

The review session will be on Friday from 3 to 4. Please meet at A401 BSA (my office).

Review: 14 November 2006

noreply@blogger.com (Murali Shanker) — Wed, 15 Nov 2006 04:30:00 +0000

We continued our review for Exam 3, and concentrated on Confidence Intervals.

Section 001
Section 002

Review: 9 November 2006

noreply@blogger.com (Murali Shanker) — Fri, 10 Nov 2006 02:47:00 +0000

We reviewed hypothesis testing today. This is part of the review for Exam 3.

Section 001
Section 002

Chapter 8: 7 November 2006

noreply@blogger.com (Murali Shanker) — Tue, 07 Nov 2006 20:07:00 +0000

We completed our discussion on hypothesis testing today.

Section 001
Section 002

Chat Tool

noreply@blogger.com (Murali Shanker) — Sun, 05 Nov 2006 23:18:00 +0000

I have now added a new tool to the blog. You can chat right here :-) Try it out.

Chapter 8: 2 November 2006

noreply@blogger.com (Murali Shanker) — Fri, 03 Nov 2006 04:55:00 +0000

Today, we started our discussion on Hypothesis Testing. We saw the three types of hypothesis, and definition of rejection region, critical values, and p-values.

Section 001
Section 002

Chapter 7: 31 October 2006

noreply@blogger.com (Murali Shanker) — Tue, 31 Oct 2006 20:18:00 +0000

We finished our discussion of confidence intervals by talking about one-sided intervals. We saw examples using both the T and the standard-normal tables.

Section 001
Section 002

Chapter 7: 26 October 2006

noreply@blogger.com (Murali Shanker) — Fri, 27 Oct 2006 16:48:00 +0000

We continued our discussion on confidence intervals, and introduced the t distribution. The t distribution is used when we don't have the population standard deviation, and instead use the sample standard deviation s. All other assumptions remain in calculating the confidence intervals.

As expected, Lecture123 in Section 2 crashed. But, I have been able to recover most of it, so it is in two parts. The easiest way is to just listen to Section 001 lecture.

Section 001
Section 002 - part 1
Section 002 - part 2

Chapter 7: 24 October 2006

noreply@blogger.com (Murali Shanker) — Wed, 25 Oct 2006 19:25:00 +0000

Today's topic was on interval estimation. Specifically, we talked about confidence intervals. We made assumptions on the distributional form, and that we knew σ. In the next class, we will relax some of these assumptions.

As with technology, things did go wrong. I completely lost Section 001 lecture. Thankfully, the Section 002 lecture worked. So, please use that to listen to your lectures.

Section 002

Review - Exam 2: 19 October 2006

noreply@blogger.com (Murali Shanker) — Thu, 19 Oct 2006 19:13:00 +0000

Exam 2:

25 questions
6-8 questions from Chapter 1-3, rest from 4-6.
4-5 interpretation type questions. Know your definitions.

Section 001
Section 002

Chapter 6: 17 October 2006

noreply@blogger.com (Murali Shanker) — Wed, 18 Oct 2006 03:45:00 +0000

We finished our discussion of sampling discussions by discussing properties of estimators.

An estimator is unbiased if on average the value of the estimator is equal to the parameter it is estimating.
An estimator is consistent if the larger the sample size, the closer is the value of the estimator to the parameter it is estimating.
An estimator is efficient, if it has the smallest variance among other unbiased estimators.

Section 001
Section 002

Additional Problems for Exam 2 - Section 001

Chapter 6: 12 October 2006

noreply@blogger.com (Murali Shanker) — Thu, 12 Oct 2006 18:35:00 +0000

We continued our discussion on sampling distributions. We saw four important points:

The average of all the sample means is equal to the population mean μ. That is, E(Xbar) = μ
The variance of the sample mean is equal to the variance of the population divided by the sample size. That is, σ²_xbar= σ²/n.
When the population is normally distributed, the sample mean distribution is also normal. That is, if X ~ N, Xbar ~ N.
When we don't know the distribution of the population, the distribution of the sample mean is approximately normally distributed for large sample sizes. This is called the central limit theorem.

Section 001
Section 002

Chapter 6: 10 October 2006

noreply@blogger.com (Murali Shanker) — Tue, 10 Oct 2006 18:44:00 +0000

Today's lecture, or at least I tried to, was on Sampling Distributions. The idea behind sampling distributions is to understand the behavior of the sample mean. By doing that, we can then be able to predict the population mean more accurately. As an exercise, I asked each group to calculate the population mean (N=8). I then asked each group to take samples of size n=7, and for each sample, calculate the sample mean. You should have observed the following results:

The average of the sample means, i.e., E(Xbar) = μ, the population mean. In the next class, we will talk about other properties of the sampling distribution of the sample mean. Today's Section 001 lecture may be a little off the chart. Those who attended know why, I hope :-)

Section 001
Section 002

Chapter 4: 5 October 2006

noreply@blogger.com (Murali Shanker) — Thu, 05 Oct 2006 18:14:00 +0000

We continued our discussion on the Normal Distribution. I have posted several additional problems for Exam 2, and also please look at the sample problems of my Wednesday, September 13, 2006 blog post. That contains more questions on the Normal Distribution. Here are the lectures:

Section 001
Section 002

Exam 1 Results

noreply@blogger.com (Murali Shanker) — Wed, 04 Oct 2006 18:11:00 +0000

Count:	337
Average:	39.5
Median:	40.0
Maximum:	50.0
Minimum:	19.3
Standard Deviation:	5.87

Glossary

noreply@blogger.com (Murali Shanker) — Wed, 04 Oct 2006 17:49:00 +0000

I have now added a link to a glossary of terms used in our class. This list is not complete, but I will continue to add terms as the semester progresses.

Chapter 4: 3 October 2006

noreply@blogger.com (Murali Shanker) — Tue, 03 Oct 2006 21:17:00 +0000

Today, we continued our discussion on random variables, and specifically, continuous random variables. We started our discussion on Normal Distribution. Specifically, we saw the properties of the Normal distribution, and how to convert to Standard Normal, and then use the tables to determine probabilities. The lectures are here:

Section 001
Section 002

I also asked you to look over Exam 1. Please give me specific examples of the types of questions you are having trouble with it, and how I can help.

Chapter 4: 28 September 2006

noreply@blogger.com (Murali Shanker) — Thu, 28 Sep 2006 18:38:00 +0000

Today, we discussed random variables, and probability distributions. We saw examples of discrete distributions, and started discussing the Normal distribution.

Section 001
Section 002

Exam 1 review: 26 September 2006

noreply@blogger.com (Murali Shanker) — Wed, 27 Sep 2006 15:47:00 +0000

We reviewed chapters 1-3. There were some techical problems in Section 001, but I still managed to capture the session, just not in Lecture123. Section 001 students can listen to Section 002, if you want to do it through Lecture123, or directly in the links given below:

Section 001
Section 002

Answers to additional problems for exam 1

noreply@blogger.com (Murali Shanker) — Sat, 23 Sep 2006 17:25:00 +0000

Q. You are given the following data: 23,34,11,40,25,47
assuming that these data reflect the population of interest, these
data can be considered symmetric.
A. Calculate the mean and median. If the values are the same, then the data are symmetric.
Q. If a set of data has 1,500 values, the 30th percentile
value will correspond to the 450th value in the data when the data
have been arranged in numerical order.
A. True. .30 x 1500 = 450. That indicates that the 450 values is the 30th percentile.
Q. A nuclear power plant produces a large amount of heat that is discharged into the water system. This heat can raise the temperature of the water system which leads to an increase in the concentration of chlorophyll-a and thus a longer growing season. To study this effect, water samples were collected monthly for one year at 3 stations and the concentration of chlorophyll-a (in milligrams per liter, mg/liter) was measured. Station 1 is closest to the source of discharge while Station 3 is furthest away. The data were used to produce the following side-by-side boxplots. What is (approximately) the largest concentration of chlorophyll-a in mg/liter for Station 2?

A. The largest concentration for Station 2 appears in March. That value is approximately 17.
Q. Suppose a study of houses that have sold recently in your community
showed the following frequency distribution for the number of bedrooms:

Bedrooms Frequency
1 1
2 18
3 140
4 57
5 11
Based on this information the mean number of bedrooms in houses that
sold is approximately 3.26. Explain?
A. The first column is the number of bedrooms, and the second column is the count or frequency, i.e., how many houses with such bedrooms. Therefore, the total number of bedrooms there are:

1x1 + 2 x18 + 3x140 + 4x57 + 5x11 = 740

You need to divide this total by the total number of houses to get the average number of bedrooms per house:

Average = 740 / (1 + 18 + 140 + 57 + 11) = 740 / 227 = 3.26

Chapter 3: 21 September 2006

noreply@blogger.com (Murali Shanker) — Fri, 22 Sep 2006 02:01:00 +0000

Today, we completed chapter 3 by talking about Standard Deviation, and Coefficient of Variation. We also discussed linear transformations, and a special case, standardization.

Lectures:

Section 001
Section 002

Additional Problems for Exam 1

Section 001
Section 002