Wk4 DQ - Data Analysis & Business Intelligence

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Two-Sample Tests of Hypothesis

Chapter 11

11-1

In this chapter, we expand the idea of hypothesis testing to two populations. We begin with the case in which we select random samples from two independent populations and investigate whether these populations have the same mean.

Learning Objectives

LO11-1 Test a hypothesis that two independent population means are equal, assuming that the population standard deviations are known and equal

LO11-2 Test a hypothesis that two independent population means are equal, with unknown population standard deviations

LO11-3 Test a hypothesis about the mean population difference between paired or dependent observations

LO11-4 Explain the difference between dependent and independent samples

11-2

Comparing Two Population Means

In comparing two populations, we wish to know whether their means could be equal

We are investigating whether the distribution of the difference between the means could have a mean of 0

Examples

Is there a difference in the mean value of residential real estate sold by male agents and female agents in south Florida?

Is there an increase in the production rate after music is piped into the production area?

11-3

We’ll select random samples from two populations to determine whether the population means are equal. Here are examples of some questions we might want to test.

Comparing Two Population Means (2 of 2)

11-4

Recall that the test statistic follows the standard normal distribution when the population standard deviations are known. We’ll use formula 11-2 in the following hypothesis test.

Comparing Two Population Means Example

Customers at the FoodTown Supermarket have a choice when paying for their groceries. They may check out and pay using the standard cashier-assisted checkout or they may use the new Fast Lane procedure (self-checkout). The store manager would like to know if the mean checkout time using the standard checkout method is longer than using the Fast Lane. The time was measured from when the customer enters the line until all his or her bags are in the cart.

11-5

Comparing Two Population Means Example (2 of 3)

Step 4: Formulate the decision rule, Reject H0 if z > 2.326

11-6

The alternate hypothesis sign points to the right, so this is a one-tailed, right-tailed test. Go to the t distribution (Appendix B.5); in the table headings, find the row labeled “Level of Significance for One-Tailed Test” and select the column for an alpha of .01. Go to the bottom row with infinite degrees of freedom. The critical z value is 2.326. We assume the population standard deviations are .40 and .30 as shown in the table; then take the sample and calculate the test statistic z = 3.123 which is greater than the critical value of 2.326. We reject the null hypothesis.

Comparing Two Population Means Example (3 of 3)

11-7

Step 5: Make the decision regarding H0. FoodTown randomly selected 50 customers using the standard checkout and computed a mean time of 5.5 minutes and randomly selected 100 customers using the Fast Lane and computed a mean time of 5.3 minutes. We will reject the null hypothesis.

Step 6: Interpret the result. The difference of .20 minute is too large to have occurred by chance. We conclude the Fast Lane method is faster.

The test statistic of 3.123 is greater than our critical value of 2.326. Therefore, our decision is to reject the null hypothesis.

Compare Two Means Using t

There are two major differences in this test and the test just described in this chapter

We assume the sampled populations have equal but unknown standard deviations

We use the t distribution

The three requirements for the test:

The sampled populations are approximately normally distributed

The sampled populations are independent

The standard deviations of the two populations are equal

11-8

The test statistic to compare two means is the t distribution if the population standard deviations are not known.

Compare Two Means Using t (2 of 2)

Finding the value of t requires two steps

The first step is to pool the standard deviations according to the following formula

The value of t is computed from the following formula

The degrees of freedom for the test are n1 + n2 − 2

11-9

Two-Sample Pooled Test Example

Owens Lawn Care Inc. manufactures and assembles lawnmowers that are shipped to dealers throughout the United States and Canada. Two different procedures have been proposed for mounting the engine on the frame of the lawnmower, the Welles method and the Atkins method. The question is, is there a difference in the methods’ mean time to mount the engines on the frames of the lawnmowers?

A time and motion study is conducted to evaluate.

11-10

Here are the results of a sample of 5 employees timed using the Welles method and six using the Atkins method. We decide to use the .10 significance level.

Two-Sample Pooled Test Example Continued

Step 4: Formulate the decision rule, do not reject H0 if t falls between −1.833 and 1.833

Step 5: Make decision regarding H0.

It takes three steps to compute the

value of t.

First:

Calculate the sample standard deviations

11-11

The degrees of freedom is 9 (nW+nA-number of samples; so, 5 + 6 − 2 = 9); the critical number values of t from Appendix B.5 for df = 9, a two-tailed test, and the .10 significance level are −1.833 and 1.833. Use formula 3-11 to calculate the sample standard deviations.

Two-Sample Pooled Test Example Concluded

Second: Pool the sample variances

Third: Determine the value of t

The decision is not to reject the null hypothesis because –0.662 falls in the region between −1.833 and 1.833.

Step 6: Interpret the result; we conclude the sample data failed to show a difference between the mean assembly times of the two methods.

11-12

Use formula 11-3 to pool the sample variances (sample standard deviations squared). Use formula 11-4 to calculate t. Excel has a procedure that will perform these calculations. The details can be found in Appendix C.

Unequal Population Standard Deviations

If we cannot assume the population standard deviations are equal, we adjust the degrees of freedom and the formula for finding t

We determine the degrees of freedom based on the following formula

The value of the test statistic is computed from the following formula

11-13

In the previous examples, although we did not know the population standard deviations, we assumed they were equal. In many cases, that is a reasonable assumption, but what if it is not? If it is not reasonable to assume the population standard deviations are equal then we use s1 and s2 in place of the population standard deviations and n1 and n2 are the respective sample sizes. We adjust the degrees of freedom downward using formula 11-6; this results in a larger value of the test statistic for rejecting the null hypothesis.

Unequal Population Standard Deviations Example

Personnel in a consumer testing laboratory are evaluating the absorbency of paper towels. They wish to compare a set of store brand towels to a similar group of name brand towels. For each brand, they dip a ply of the paper into a tub of fluid, allow the paper to drain back into the vat for 2 minutes, and then evaluate the amount of liquid the paper has taken up from the vat.

A random sample of 9 store brand towels absorption amounts (in ml.)

A random sample of 12 name brand towels absorption amounts (in ml.)

11-14

After observing there is more variation in the store brand, it does not seem reasonable to assume the population standard deviations are equal. So we decide to use the t distribution and assume the population standard deviations are not the same.

Unequal Population Standard Deviations Example (2 of 3)

We must adjust the degrees of freedom with formula 11-6 before finding the critical values and round the result down to an integer; in this case, 10

11-15

After adjusting the degrees of freedom with formula 11-6, we use Appendix B.5 to find the critical values; with 10 degrees of freedom, a two-tailed test, and the .10 significance level, the critical t values are −1.812 and 1.812. We use formula 11-5 to find the test statistic; −2.474; which falls in the rejection region, so our decision is to reject the null hypothesis. We conclude the mean absorption rate for the two towels is not the same.

Unequal Population Standard Deviations Example (3 of 3)

11-16

Step 4: State the decision rule, do not reject H0 if t falls between −1.812 and 1.812.

Step 5: Make decision

Step 6: Interpret; the mean absorption rate of the two types of towels is not the same.

The decision is to reject the null hypothesis because –2.474 is smaller than −1.812.

Dependent Samples

11-17

For dependent samples, we assume the population distribution of the paired differences has a mean of 0. See the upcoming slides for an example.

Dependent Samples Continued

Note: the standard deviation of the differences will be computed with the formula 3-11, except d is substituted for x

Example

Nickel Savings and Loan employs two firms, Schadek Appraisals and Bowyer Real Estate, to appraise the value of the real estate on which it makes loans. To review the consistency of the two appraisal firms, Nickel randomly selects 10 homes and has both of the firms appraise the values of the selected homes. Thus, there will be a pair of values for each home; these appraised values are related to the home selected. This is called a paired sample.

11-18

If the two appraisal firms are reporting similar estimates, then sometimes Schadek Appraisals will be the higher value and sometimes Bowyer Real Estate will have the higher value. However, the mean of the distribution of differences will be 0. On the other hand, if one of the firms consistently reports larger appraisal values, then the mean of the distribution of the differences will not be 0.

Dependent Samples Example

Recall that Nickel Savings and Loan wishes to compare the two companies it uses to appraise the value of residential homes. Nickel Savings selected a sample of 10 residential properties and scheduled both firms for an appraisal. The results are reported in $000. At the .05 significance level, can we conclude there is a difference between the firm’s appraised values?

11-19

If the population mean of the differences is 0, then we conclude there is no difference between the two firms’ appraised values. A two-tailed test is appropriate because we are interested in whether there is a difference in the firms’ appraised values. Using Appendix B.5, we have a two-tailed test, 9 degrees of freedom (n-1), and a level of significance of .05, so the critical values are plus or minus 2.262.

Dependent Samples Example Continued

Step 5: Make your decision; we’ll reject the null hypothesis

Step 6: Interpret; we conclude there is a difference between the firms’ mean appraised home values

11-20

First, compute the mean (4.6) and the standard deviation (4.402) of the sample differences. Then compute the test statistic, t (3.305). We decide to reject the null hypothesis; the test statistic t, is greater than the critical value, 2.262. The two firms’ appraised property values are different; in other words, each firm is appraising the same property with a different value.

Excel statistical software can perform the calculations in formula 11-7 for us. See your textbook for details.

Dependent and Independent Samples

There are two types of dependent samples

Those characterized by a measurement, an intervention of some type, and then another measurement

For example, suppose we wish to show that by playing music in the production area we are able to increase production. We begin by selecting a sample of workers and measure their output, then we place the speakers in the production area and play soothing music, and then we again measure the output

A matching or pairing of the observations

For example, the Nickel Savings and Loan example illustrates dependent samples because a property is selected and both firms appraise the same property

We prefer a test based on dependent samples because it reduces the amount of variation in the test and is considered a better test

11-21

The first type of dependent sample is sometimes referred to as the “before” and “after” study. In the Nickel S & L example, what if we had assumed that we have two independent samples rather than dependent samples of real estate? We would have pooled the variances of the samples and then used formula 11-4 for t, t would equal .716 which is much less than the critical value 2.101. The null hypothesis would not be rejected.

The hypothesis test based on dependent samples eliminates the variation between the values of the properties and focuses only on the comparisons in the two appraisals for each property.

Chapter 11 Practice Problems

11-22

Question 1

11-23

A sample of 40 observations is selected from one population with a population standard deviation of 5. The sample mean is 102. A sample of 50 observations is selected from a second population with a population standard deviation of 6. The sample mean is 99. Conduct the following test of hypothesis using the .04 significance level.

H0: μ1 = μ2

H1: μ1 ≠ μ2

Is this a one-tailed or a two-tailed test?

State the decision rule.

Compute the value of the test statistic.

What is your decision regarding H0?

What is the p-value?

LO11-1

Question 7

11-24

A random sample of 10 observations from one population revealed a sample mean of 23 and a sample standard deviation of 4. A random sample of 8 observations from another population revealed a sample mean of 26 and a sample standard deviation of 5. At the .05 significance level, is there a difference between the population means?

The null and alternate hypotheses are:

H0: μ1 = μ2

H1: μ1 ≠ μ2

State the decision rule

Compute the pooled estimate of the population variance

Compute the test statistic

State your decision about the null hypothesis

Estimate the p-value.

LO11-2

Question 15

11-25

A recent survey compared the costs of adoption through public and private agencies. For a sample of 16 adoptions through a public agency, the mean cost was $21,045, with a standard deviation of $835. For a sample of 18 adoptions through a private agency, the mean cost was $22,840, with a standard deviation of $1,545. Assume the sample populations do not have equal standard deviations and use the .05 significance level

Determine the number of degrees of freedom and round down to the nearest integer value

State the decision rule

Compute the value of the test statistic

State your decision about the null hypothesis

LO11-2

Question 17

11-26

The null and alternate hypotheses are:

H0: μd ≤ 0

H1: μd > 0

The following sample information shows the number of defective units produced on the day shift and the afternoon shift for a sample of 4 days last month.

What is the p-value?

Is the null hypothesis rejected?

What is the conclusion indicated by the analysis?

LO11-3