Up to this point, we have discussed inferences regarding a single population parameter (e.g., μ, p, σ2). We have used sample data to construct confidence intervals to estimate the population mean or proportion and to test hypotheses about the population mean and proportion. In both of these chapters, all the examples involved the use of one sample to form an inference about one population. Frequently, we need to compare two sets of data, and make inferences about two populations. This chapter deals with inferences about two means, proportions, or variances. For example:

• You are studying turkey habitat and want to see if the mean number of brood hens is different in New York compared to Pennsylvania.
• You want to determine if the treatment used in Skaneateles Lake has reduced the number of milfoil plants over the last three years.
• Is the proportion of people who support alternative energy in California greater compared to New York?
• Is the variability in application different between two mist blowers?

These questions can be answered by comparing the differences of:

• Mean number of hens in NY to the mean number of hens in PA.
• Number of plants in 2007 to the number of plants in 2010.
• Proportion of people in CA to the proportion of people in NY.
• Variances between the mist blowers.

This chapter is comprised of five sections. The first and second sections examine inferences about two means with two independent samples. The third section examines inferences about means with two dependent samples, the fourth section examines inferences about two proportions, and the fifth section examines inferences between two variances.

Section 1

Inferences about Two Means with Independent Samples (Assuming Unequal Variances)

Using independent samples means that there is no relationship between the groups. The values in one sample have no association with the values in the other sample. For example, we want to see if the mean life span for hummingbirds in South Carolina is different from the mean life span in North Carolina. These populations are not related, and the samples are independent. We look at the difference of the independent means.

In Chapter 3, we did a one-sample t-test where we compared the sample mean () to the hypothesized mean (μ). We expect that would be close to μ. We use the sample mean, the sample standard deviation, and the sample size for the one-sample test.

With a two-sample t-test, we compare the population means to each other and again look at the difference. We expect that would be close to μ1 – μ2. The test statistic will use both sample means, sample standard deviations, and sample sizes for the test.

• For a one-sample t-test we used as a measure of the standard deviation (the standard error).
• We can rewrite →.
• The numerator of the test statistic will be
• This has a standard deviation of .

A two-sample t-test follows the same four steps we saw in Chapter 3.

• Write the null and alternative hypotheses.
• State the level of significance and find the critical value. The critical value, from the student’s t-distribution, has the lesser of n1-1 and n2 -1 degrees of freedom.
• Compute the test statistic.
• Compare the test statistic to the critical value and state a conclusion.

The assumptions we saw in Chapter 3 still must be met. Both samples come from independent random samples. The populations must be normally distributed, or both have large enough sample sizes (n1 and n2 ≥ 30). We will also use the same three pairs of null and alternative hypotheses.

Table 1. Null and alternative hypotheses.

Rewriting the null hypothesis of μ1 = μ2 to μ1 – μ2 = 0, simplifies the numerator. The test statistic is Welch’s approximation (Satterthwaite Adjustment) under the assumption that the independent population variances are not equal.

This test statistic follows the student’s t-distribution with the degrees of freedom adjusted by

A simpler alternative to determining degrees of freedom when working a problem long-hand is to use the lesser of n1-1 or n2-1 as the degrees of freedom. This method results in a smaller value for degrees of freedom and therefore a larger critical value. This makes the test more conservative, requiring more evidence to reject the null hypothesis.

Example 1

A forester is studying the number of cavity trees in old growth stands in Adirondack Park in northern New York. He wants to know if there is a significant difference between the mean number of cavity trees in the Adirondack Park and the old growth stands in the Monongahela National Forest. He collects two independent random samples from each forest. Use a 5% level of significance to test this claim.

Adirondack Park	Monongahela Forest
n1 = 51 stands	n2 = 56 stands
= 39.6	= 43.9
s1 = 9.4	s2 = 10.7

1) H0: μ1 = μ2 or μ1 – μ2 = 0 There is no difference between the two population means.

H1: μ1 ≠ μ2 There is a difference between the two population means.

2) The level of significance is 5%. This is a two-sided test so alpha is split into two sides. Computing degrees of freedom using the equation above gives 105 degrees of freedom.

The critical value (), based on 100 degrees of freedom (closest value in the t-table), is ±1.984. Using 50 degrees of freedom, the critical value is ±2.009.

3) The test statistic is

4) The test statistic falls in the rejection zone.

Figure 1. A comparison of the critical values and test statistic.

We reject the null hypothesis. We have enough evidence to support the claim that there is a difference in the mean number of cavity trees between the Adirondack Park and the Monongahela National Forest.

Construct and Interpret a Confidence Interval about the Difference of Two Independent Means

A hypothesis test will answer the question about the difference of the means. BUT, we can answer the same question by constructing a confidence interval about the difference of the means. This process is just like the confidence intervals from Chapter 2.

1. Find the critical value.
2. Compute the margin of error.
3. Point estimate ± margin of error.

Because we are working with two samples, we must modify the components of the confidence interval to incorporate the information from the two populations.

• The point estimate is .
• The standard error comes from the test statistic
• The critical value comes from the student’s t-table.

The confidence interval takes the form of the point estimate plus or minus the standard error of the differences.

±

We will use the same three steps to construct a confidence interval about the difference of the means.

1. critical value
2. E =
3. ± E

Confidence Interval Interpretations

• If the confidence interval contains all positive values, we find a significant difference between the groups, AND we can conclude that the mean of the first group is significantly greater than the mean of the second group.
• If the confidence interval contains all negative values, we find a significant difference between the groups, AND we can conclude that the mean of the first group is significantly less than the mean of the second group.
• If the confidence interval contains zero (it goes from negative to positive values), we find NO significant difference between the groups.

In this problem, the confidence interval is (-8.204, -0.396). We have all negative values, so we can conclude that there is a significant difference in the mean number of cavity trees AND that the mean number of cavity trees in the Adirondack forests is significantly less than the mean number of cavity trees in the Monongahela Forest. The confidence interval gives an estimate of the mean difference in number of cavity trees between the two forests. There are, on average, 0.396 to 8.204 fewer cavity trees in the Adirondack Park than the Monongahela Forest.

P-value Approach

We can also use the p-value approach to answer the question. Remember, the p-value is the area under the normal curve associated with the test statistic. This example is a two-sided test (H1: μ1 ≠ μ2 ) so the p-value, when computed by hand, will be multiplied by two.

The test statistic equals -2.213, so the p-value is two times the area to the left of -2.213. We can only estimate the p-value using the student’s t-table. Using the lesser of n1– 1 or n2– 1 as the degrees of freedom, we have 50 degrees of freedom. Go across the 50 row in the student’s t-table until you find the absolute value of the test statistic. In this case, 2.213 falls between 2.109 and 2.403. Going up to the top of each of those columns gives you the estimate of the p-value (between 0.02 and 0.01).

Table 2. Student t-Distribution

The p-value is 2x(0.01 – 0.02) = (0.02 < p < 0.04). The p-value is greater than 0.02 but less than 0.04. This is less than the level of significance (0.05), so we reject the null hypothesis. There is enough evidence to support the claim that there is a significant difference in the mean number of cavity trees between the areas.

Example 2

Researchers are studying the relationship between logging activities in the northern forests and amphibian habitats. They were comparing moisture levels between old-growth and post-harvest habitats. The researchers believe that post-harvest habitat has a lower moisture level. They collected data on moisture levels from two independent random samples. Test their claim using a 5% level of significance.

Old Growth	Post Harvest
n1 = 26	n2 = 31
=0.62 g/cm3	= 0.56 g/cm3
s1 = 0.12 g/cm3	s2 = 0.17 g/cm3

H0: μ1 = μ2 or μ1 – μ2 = 0. There is no difference between the two population means.

H1: μ1 > μ2. Mean moisture level in old growth forests is greater than post-harvest levels.

We will use the critical value based on the lesser of n1– 1 or n2– 1 degrees of freedom. In this problem, there are 25 degrees of freedom and the critical value is 1.708. Now compute the test statistic.

The test statistic does not fall in the rejection zone. We fail to reject the null hypothesis. There is not enough evidence to support the claim that the moisture level is significantly lower in the post-harvest habitat.

Now answer this question by constructing a 90% confidence interval about the difference of the means.

1) = 1.708

2) E =

3) ± E (0.62-0.56) ±0.0658

The 90% confidence interval for the difference of the means is (-0.0058, 0.1258). The values in the confidence interval run from negative to positive indicating that there is no significant different in the mean moisture levels between old growth and post-harvest stands.

Software Solutions

Minitab

The p-value (0.064) is greater than the level of confidence so we fail to reject the null hypothesis.

Additional example: www.youtube.com/watch?v=7pIb-GVixFo.

Excel

The one-tail p-value (0.063809) is greater than the level of significance, therefore, we fail to reject the null hypothesis.

Section 2

Pooled Two-sampled t-test (Assuming Equal Variances)

In the previous section, we made the assumption of unequal variances between our two populations. Welch’s t-test statistic does not assume that the population variances are equal and can be used whether the population variances are equal or not. The test that assumes equal population variances is referred to as the pooled t-test. Pooling refers to finding a weighted average of the two independent sample variances.

The pooled test statistic uses a weighted average of the two sample variances.

If n1= n2, then S2p = (1/2)s21 + (1/2)s22, the average of the two sample variances. But whenever n1≠n2, the s2 based on the larger sample size will receive more weight than the other s2.

The advantage of this test statistic is that it exactly follows the student’s t-distribution with n1+ n2– 2 degrees of freedom.

The hypothesis test procedure will follow the same steps as the previous section.

It may be difficult to verify that two population variances might be equal based on sample data. The F-test is commonly used to test variances but is not robust. Small departures from normality greatly impact the outcome making the results of the F-test unreliable. It can be difficult to decide if a significant outcome from an F-test is due to the differences in variances or non-normality. Because of this, many researchers rely on Welch’s t when comparing two means.

Software Solutions

Minitab

Two-Sample T-Test and CI: Substrate1, Substrate2

The p-value (0.000) is less than the level of significance (0.05). We will reject the null hypothesis.

Excel

This is a one-sided test (greater than) so use the P(T<=t) one-tail value 2.43E-05. The p-value (0.0000243) is less than the level of significance (0.05). We will reject the null hypothesis.

Section 3

Inferences about Two Means with Dependent Samples—Matched Pairs

Dependent samples occur when there is a relationship between the samples. The data consists of matched pairs from random samples. A sampling method is dependent when the values selected for one sample are used to determine the values in the second sample. Before and after measurements on a population, such as people, lakes, or animals are an example of dependent samples. The objects in your sample are measured twice; measurements are taken at a certain point in time, and then re-taken at a later date. Dependency also occurs when the objects are related, such as eyes or tires on a car. Pairing isn’t a problem; it’s an opportunity to use the information that occurs with both measurements.

Before you begin your work, you must decide if your samples are dependent. If they are, you can take advantage of this fact. You can use this matching to better answer your research questions. Pairing data reduces measurement variability, which increases the accuracy of our statistical conclusions.

We use the difference (the subtraction) of the pairs of data in our analysis. For each pair, we subtract the values:

• d1 = before1 – after 1
• d2 = before 2 – after 2
• d3 = before 3 – after 3
• …

We are creating a new random variable d (differences), and it is important to keep the sign, whether positive or negative. We can compute d̄, the sample mean of the differences, and sd, the sample standard deviation of the differences as follows:

Just as we used the sample mean and the sample standard deviation in a one-sample t-test, we will use the sample mean and sample standard deviation of the differences to test for matched pairs. The assumption of normality must still be verified. The differences must be normally distributed or the sample size must be large enough (n ≥ 30).

We can do a hypothesis test using matched pairs data following the same methods we used in the previous chapter.

• Write the null and alternative hypotheses.
• State the level of significance and find the critical value.
• Compute a test statistic.
• Compare the test statistic to the critical value and state a conclusion.

Since we are using the differences between the pairs of data, we identify this in our null and alternative hypotheses: H0: μd = 0. The mean of the differences is equal to zero; there is no difference in “before and after” values.

We’ll use the same three pairs of null and alternative hypotheses we used in the previous chapter.

Table 3. Null and alternative hypotheses.

The critical value comes from the student’s t-distribution table with n – 1 degrees of freedom, where n = number of matched pairs. The test statistic follows the student’s t-distribution

The conclusion must always answer the question you are asking in the alternative hypothesis.

• Reject the H0. There is enough evidence to support the alternative claim.
• Fail to reject the H0. There is not enough evidence to support the alternative claim.

Example 4

An environmental biologist wants to know if the water clarity in Owasco Lake is improving. Using a Secchi disk, she takes measurements in specific locations at specific dates during the course of the year. She then repeats the measurements in the same locations and on the same dates five years later. She obtains the following results:

Date	Initial Depth	5-year Depth	Difference
5/11	38	52	-14
6/7	58	60	-2
6/24	65	72	-7
7/8	74	72	2
7/27	56	54	2
8/31	36	48	-12
9/30	56	58	-2
10/12	52	60	-8

Using a 5% level of significance, test the biologist’s claim that water clarity is improving.

The data are paired by date with two measurements taken at each point five years apart. We will use the differences (right column) to see if there has been a significant improvement in water clarity. Using your calculator, Minitab, or Excel, compute the descriptive statistics on the differences to get the sample mean and the sample standard deviation of the differences.

1) The null and alternative hypotheses:

Ho: μd = 0 (The mean of the differences is equal to zero- no difference in water clarity over time.)

H1: μd < 0 (The water clarity is improving.)

We test “less than” because of how we computed the differences and the question we are asking.

In this case, we hope to see greater depth (better water clarity) at the five-year measurements. By calculating Initial – 5-year we hope to see negative values, values less than zero, indicating greater depth and clarity at the 5-year mark. Think of it like this:

Initial Depth < 5-year depth

This gives you the direction of the test!

2) The critical value tα.

The critical value comes from the student’s t-distribution table with n – 1 degrees of freedom. In this problem, we have eight pairs of data (n = 8) with 7 degrees of freedom. This is a one-sided test (less than), so alpha is all in the left tail. Go down the 0.05 column with 7 df to find the correct critical value (tα) of -1.895.

3) The test statistic = .

We subtract zero from d-bar because of our null hypothesis. Our null hypothesis is that the difference of the before and after values are statistically equal to zero. In other words, there has been no change in water clarity.

4) Compare the test statistic to the critical value and state a conclusion.

The test statistic (-2.38) is less than the critical value (-1.895). It falls in the rejection zone.

Figure 2. Comparison of the critical value and the test statistic.

We reject the null hypothesis. We have enough evidence to support the claim that the mean water clarity has improved.

P-value Approach

We can also use the p-value approach to answer the question. To estimate p-value using the student’s t-table, go across the row for 7 degrees of freedom until you find the two values that the absolute value of your test statistic falls between.

Table 4. Student t-Distribution.

The p-value for this test statistic is greater than 0.02 and just less than 0.025. Compare this to the level of significance (alpha). The Decision Rule says that if the p-value is less than α, reject the null hypothesis. In this case, the p-value estimate (0.02 – 0.025) is less than the level of significance (0.05). Reject the null hypothesis. We have enough evidence to support the claim that the mean water clarity has improved.

BUT, what if you used a 1% level of significance? In this case, the p-value is NOT less than the level of significance ((0.02 – 0.025)>0.01). We would fail to reject the null hypothesis. There is NOT enough evidence to support the claim that the water clarity has improved. It is important to set the level of significance at the start of your research and report the p-value. Another researcher may interpret your findings differently, based on your reported p-value and their own selected level of significance.

Construct and Interpret a Confidence Interval about the Differences of the Data for Matched Pairs