Math 365, Elementary Statistics

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Lesson 8 : Comparing Two Populations

Introduction

In this lesson we try to compare two populations. We will consider the following:

1. Compute a confidence interval of the difference μ₁- μ₂ of the means of two populations. For example, we may like to estimate the difference μ₁ - μ₂ between the mean μ₁ = annual male income and the mean μ₂ = annual female income in the United States.
2. Compute a confidence interval of the difference p₁-p₂ of the proportions of an attribute present (or proportions of "success") in two populations. For example, we may like to estimate the difference p₁-p₂ between p₁ = the proportion of defective items produced by the new machine and p₂ = the proportion of defective items produced by the old machine.

8.1 Confidence Interval of μ₁- μ₂

Suppose X, Y are two similar random variables. Let mean and standard deviation of X be, respectively, μ₁ and σ₁. Let mean and standard deviation of Y be, respectively, μ₂ and σ₂. We want to compute a confidence interval for the difference μ₁- μ₂. So we do the following.

1. We draw a sample X₁, X₂, …, X_m, of size m, from the X population and we draw a sample Y₁, Y₂, …, Y_n, of size n, from the Y population. Let

   X  = (X₁+X₂+ … +X_m)/m
   
   Y  = (Y₁+Y₂+ … +Y_n)/n

   be the corresponding sample means.

2. BY CLT, we have that X has

   N(μ₁, σ₁/√m )

   distribution and Y has

   N(μ₂, σ₂/√n )

   distribution.

3. You would agree that X-Y is a natural estimator of μ₁- μ₂.
4. Now we assume that the X samples and Y samples are mutually independent. In that case, it follows that X-Y has

   N(μ₁ - μ₂, σ) - distribution,
   
   where
   
   σ =  √( σ₁²/m + σ₂²/n ).

5. It follows that

   P(-z_α/2 < ((X-Y) - (μ₁ - μ₂)) /σ < z_α/2 )  =   1 - α.
   
   where σ is as above in (4).

6. If we simplify, we get

   P(X-Y -z_α/2 σ < μ₁ - μ₂ < X-Y +z_α/2 σ )  =   1 - α.
   
   where σ is as above in (4).

7. Theorem. A (1-α)100 percent confidence interval for μ₁- μ₂ is given by
   
   
   x-y -z_α/2 σ    <    μ₁ - μ₂    <    x-y +z_α/2 σ
   
   where σ is as above in (4).

   This formula is usable if we know the values σ₁ and σ₂.

8. The margin of error is given by
   

   E  =  z_α/2 σ
   
   where σ is as above in (4).

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Use of Calculators (if you have a TI-83): 2-SampZinterval
Press stat and then select TESTS. Select 2-SampZinterval and enter. Input: you will have to select stats (not data) in this section. Feed in the values of σ1, σ2, x,y, m, n and c-level. Select calculate and enter. It will give you the confidence interval. The margin of error = E = (width of the interval)/2.

Problems on 8.1: Confidence Interval of μ₁ - μ₂

Exercise 8.1.1. Suppose we have two normal populations with means μ₁, μ₂ and standard deviation σ₁, σ₂ respectively. It is known that σ₁ = 8.1 and σ₂ = 11.3. A sample of size m = 64 was collected from the first population, and the sample mean was found to be x = 3.7. A sample of size n = 99 was collected from the second population, and the sample mean was found to be y = 4.1. Compute a 95 percent confidence interval for the difference of mean μ₁- μ₂.
Solution

Exercise 8.1.2. The birth weight of babies in developed and developing countries are normally distributed with mean μ₁, μ₂ and standard deviation σ₁, σ₂, respectively. (My data is not real.) Given σ₁ = 2.3 pounds and σ₂ = 2.9 pounds. A sample of size m = 35 babies from the developed nations were collected and the sample mean birth weight was found to be x = 8.9 pounds. A sample of size n = 48 babies from the developing nations was collected and the sample mean birth weight was found to be y = 7.1 pounds.

1. Compute a point estimate of the difference of mean birth weight μ₁- μ₂.
2. Determine the margin of error of the difference μ₁- μ₂ at the 95 percent level of confidence.
3. Construct a 95 percent confidence interval for μ₁- μ₂.

Solution

Exercise 8.1.3. African elephants and Indian elephants are different in height, weight, and length of ear and tusk. It is natural to assume that all these are normally distributed. The mean height and standard deviation of African elephants are μ₁, σ₁ = 1.2 feet, respectively. The mean height and standard deviation of Indian elephants are μ₂, σ₂ = 1.1 feet, respectively. A sample of size 25 African elephants were collected and the sample mean height was found to be x = 10.9 feet. A sample of size 28 Indian elephants was collected and the sample mean height was found to be y = 9.1 feet.

1. Compute a point estimate of the difference of mean height μ₁- μ₂.
2. Determine the maximum error of the difference μ₁- μ₂ at the 99 percent level of confidence.
3. Construct a 99 percent confidence interval for μ₁- μ₂.

Solution

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8.2 When σ₁ and σ₂ are Unknown

As in the last section, we have two populations X, Y. We assume that X has N(μ₁, σ₁) distribution and Y has N(μ₂, σ₂) distribution. Unlike in the last section, we assume that σ₁, σ₂ are unknown. We try to find a confidence interval for μ₁ - μ₂.

We take a sample X₁, X₂, …, X_m of size m from the X population, and we take a sample Y₁,Y₂, …, Y_n from the Y population. Following are some facts and notations.

1. Assumptions: We make an important assumption that the variances σ₁² and σ₂² are equal. So, we write

   σ₁   =  σ₂   =  σ.

   And, we also assume that the X-sample and the Y-sample are mutually independent.

2. 	2
   Let X and S
   	X

   be the sample mean and sample variance of the X-sample and let Y and S_Y² be the sample mean and sample variance of the Y-sample.
   
   
3. Definition. Define the pooled estimate S_p² for σ² as follows

   S_p²  =
   
   [(m-1)S_X²+(n-1)S_Y² ]/ [m+n-2]  =
   
   
   [ ∑ (X_i-X)² + ∑ (Y_j-Y )² ] / [m+n-2]
   

   Although both S_X², S_Y² are estimators of σ², S_p² is a better estimator for σ² because it uses both the samples. One can see that S_p² is a weighted average of S_X² and S_Y².

4. It follows that

   T  =  [ (X - Y) - (μ₁ -μ₂) ] / [S_p√(1/m +  1/n) ]

   has a t-distribution with m+n-2 degrees of freedom.

5. Using the same kind of computations that we have done before, we see that a (1-α)100 percent confidence interval for μ₁- μ₂ is given by
   

   x-y-E   <  μ₁- μ₂   <   x-y+E

   where
   

   E=t_m+n-2,α/2 S_p √(1/m + 1/n)

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Use of Calculators (if you have a TI-83): 2-SampTint

If we have raw data, enter the data into the calculator in 2 lists (say L1,L2). Press stat and then select TESTS. Select 2-SampTinterval and enter. Input: you will have to select stats or data, depending on what is given. Feed in the values of sample standard deviation s1, s2, x, y, m, n or the Lists where you have the data and c-level. Select calculate and enter. It will give you the confidence interval and also the pooled estimate of the equal standard deviation σ. The margin of error = E = (width of the interval)/2.

Problems on 8.2: When σ₁ and σ₂ Are Unknown

Exercise 8.2.1. Suppose that we are comparing two "similar" normal populations with means μ₁, μ₂ respectively and the populations both have standard deviation σ. We collected a sample of size m = 11 from the first population that produced a sample mean x = 13.2 and sample standard deviation s₁ = 2.33. A sample of size n = 13 was collected from the second population that had sample mean y = 11.5 and sample variance s₂ = 2.73.

1. Compute the pulled estimate s_p for σ.
2. Find a point estimate for μ₁- μ₂.
3. Compute the margin of error in estimating μ₁- μ₂ at the 90 percent level of significance.
4. Compute a 90 percent confidence interval for μ₁- μ₂.

Solution

Exercise 8.2.2. Suppose we have two normal populations with means μ₁, μ₂ and equal standard deviation σ. A sample of size m = 64 was collected from the first population and the sample mean and standard deviation were found to be x = 3.7, s₁ = 9.2 . A sample of size n = 99 was collected from the second population and the sample mean and standard deviation were y = 4.1, s₂ = 8.7.

1. Compute the pulled estimate s_p for σ.
2. Compute the margin of error for a 95 percent confidence interval for μ₁- μ₂.
3. Compute a 95 percent confidence interval for the difference of mean μ₁- μ₂.

Solution

Exercise 8.2.3. The birth weight of the babies in developed and developing countries are normally distributed with mean μ₁, μ₂ and equal standard deviation σ. (My data is not real.) Suppose the following data about the birth weight from developed and developing nations were collected.

Developed 8.8 8.1 6.3 9.7 6.3 7.1 5.3 7.7 9.1 8.1 8.2 7.9 8.3 8.9 9.0 10.1 9.9 8.8 7.8 5.2 7.2

Developing 6.3 5.2 8.3 5.9 5.5 7.1 8.1 7.9 6.3 6.9 9.1 8.1 7.0 4.9 5.3 6.3 7.1 6.3 6.1 5.8 5.7 6.8 8.3 7.7

1. Compute a point estimate of the difference of mean birth weight μ₁- μ₂.
2. Compute the pulled estimate for σ.
3. Determine the maximum error of the difference μ₁- μ₂ at the 95 percent level of confidence.
4. Construct a 95 percent confidence interval for μ₁- μ₂.

Solution

Exercise 8.2.4. African elephants and Indian elephants are different in height, weight, and length of ear and tusk. It is natural to assume that all these are normally distributed. Assume that the height of African and Indian elephants have an equal mean σ. The mean heights of African elephants and Indian elephants are μ₁, μ₂, respectively. Suppose the following data were collected on the height of elephants from the two continents (these are not real data).

African 10.9 11.7 9.3 9.9 11.5 8.8 12.9 11.7 9.1 11.1 9.1 8.7 10.5 11.3 12.3 13.1 12.9 9.5 10.7 11.3

Indian 7.1 8.3 8.2 9.1 10.3 9.3 9.7 8.9 8.8 9.1 7.9 9.9 9.2 8.8 8.1 8.7 8.8 9.3 10. 1 9.9 9.9