7.3 - Comparing Two Population Means
7.3 - Comparing Two Population MeansIntroduction
In this section, we are going to approach constructing the confidence interval and developing the hypothesis test similarly to how we approached those of the difference in two proportions.
There are a few extra steps we need to take, however. First, we need to consider whether the two populations are independent. When considering the sample mean, there were two parameters we had to consider, \(\mu\) the population mean, and \(\sigma\) the population standard deviation. Therefore, the second step is to determine if we are in a situation where the population standard deviations are the same or if they are different.
Independent and Dependent Samples
It is important to be able to distinguish between an independent sample or a dependent sample.
- Independent sample
- The samples from two populations are independent if the samples selected from one of the populations has no relationship with the samples selected from the other population.
- Dependent sample
- The samples are dependent (also called paired data) if each measurement in one sample is matched or paired with a particular measurement in the other sample. Another way to consider this is how many measurements are taken off of each subject. If only one measurement, then independent; if two measurements, then paired. Exceptions are in familial situations such as in a study of spouses or twins. In such cases, the data is almost always treated as paired data.
The following are examples to illustrate the two types of samples.
Example 7-3: Gas Mileage
We want to compare the gas mileage of two brands of gasoline. Describe how to design a study involving...
- independent sample
Answer: Randomly assign 12 cars to use Brand A and another 12 cars to use Brand B.
- dependent samples
Answer: Using 12 cars, have each car use Brand A and Brand B. Compare the differences in mileage for each car.
Try it!
- We want to compare whether people give a higher taste rating to Coke or Pepsi. To avoid a possible psychological effect, the subjects should taste the drinks blind (i.e., they don't know the identity of the drink). Describe how to design a study involving independent sample and dependent samples.
- Design involving independent samples
- Design involving dependent samples
- Answer: Randomly assign half of the subjects to taste Coke and the other half to taste Pepsi.
-
Answer: Allow all the subjects to rate both Coke and Pepsi. The drinks should be given in random order. The same subject's ratings of the Coke and the Pepsi form a paired data set.
- Compare the time that males and females spend watching TV.
- We randomly select 20 males and 20 females and compare the average time they spend watching TV. Is this an independent sample or paired sample?
- We randomly select 20 couples and compare the time the husbands and wives spend watching TV. Is this an independent sample or paired sample?
- Answer: Independent Sample
-
Answer: Paired sample
The two types of samples require a different theory to construct a confidence interval and develop a hypothesis test. We consider each case separately, beginning with independent samples.
7.3.1 - Inference for Independent Means
7.3.1 - Inference for Independent MeansTwo-Cases for Independent Means
As with comparing two population proportions, when we compare two population means from independent populations, the interest is in the difference of the two means. In other words, if \(\mu_1\) is the population mean from population 1 and \(\mu_2\) is the population mean from population 2, then the difference is \(\mu_1-\mu_2\). If \(\mu_1-\mu_2=0\) then there is no difference between the two population parameters.
If each population is normal, then the sampling distribution of \(\bar{x}_i\) is normal with mean \(\mu_i\), standard error \(\dfrac{\sigma_i}{\sqrt{n_i}}\), and the estimated standard error \(\dfrac{s_i}{\sqrt{n_i}}\), for \(i=1, 2\).
Using the Central Limit Theorem, if the population is not normal, then with a large sample, the sampling distribution is approximately normal.
The theorem presented in this Lesson says that if either of the above are true, then \(\bar{x}_1-\bar{x}_2\) is approximately normal with mean \(\mu_1-\mu_2\), and standard error \(\sqrt{\dfrac{\sigma^2_1}{n_1}+\dfrac{\sigma^2_2}{n_2}}\).
However, in most cases, \(\sigma_1\) and \(\sigma_2\) are unknown, and they have to be estimated. It seems natural to estimate \(\sigma_1\) by \(s_1\) and \(\sigma_2\) by \(s_2\). When the sample sizes are small, the estimates may not be that accurate and one may get a better estimate for the common standard deviation by pooling the data from both populations if the standard deviations for the two populations are not that different.
Given this, there are two options for estimating the variances for the independent samples:
- Using pooled variances
- Using unpooled (or unequal) variances
When to use which? When we are reasonably sure that the two populations have nearly equal variances, then we use the pooled variances test. Otherwise, we use the unpooled (or separate) variance test.
7.3.1.1 - Pooled Variances
7.3.1.1 - Pooled VariancesConfidence Intervals for \(\boldsymbol{\mu_1-\mu_2}\): Pooled Variances
When we have good reason to believe that the variance for population 1 is equal to that of population 2, we can estimate the common variance by pooling information from samples from population 1 and population 2.
An informal check for this is to compare the ratio of the two sample standard deviations. If the two are equal, the ratio would be 1, i.e. \(\frac{s_1}{s_2}=1\). However, since these are samples and therefore involve error, we cannot expect the ratio to be exactly 1. When the sample sizes are nearly equal (admittedly "nearly equal" is somewhat ambiguous, so often if sample sizes are small one requires they be equal), then a good Rule of Thumb to use is to see if the ratio falls from 0.5 to 2. That is, neither sample standard deviation is more than twice the other.
If this rule of thumb is satisfied, we can assume the variances are equal. Later in this lesson, we will examine a more formal test for equality of variances.
- Let \(n_1\) be the sample size from population 1 and let \(s_1\) be the sample standard deviation of population 1.
- Let \(n_2\) be the sample size from population 2 and \(s_2\) be the sample standard deviation of population 2.
Then the common standard deviation can be estimated by the pooled standard deviation:
\(s_p=\sqrt{\dfrac{(n_1-1)s_1^2+(n_2-1)s^2_2}{n_1+n_2-2}}\)
If we can assume the populations are independent, that each population is normal or has a large sample size, and that the population variances are the same, then it can be shown that...
\(t=\dfrac{\bar{x}_1-\bar{x_2}-0}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}\)
follows a t-distribution with \(n_1+n_2-2\) degrees of freedom.
Now, we can construct a confidence interval for the difference of two means, \(\mu_1-\mu_2\).
- \(\boldsymbol{(1-\alpha)100\%}\) Confidence interval for \(\boldsymbol{\mu_1-\mu_2}\) for Pooled Variances
- \(\bar{x}_1-\bar{x}_2\pm t_{\alpha/2}s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}\)
-
where \(t_{\alpha/2}\) comes from a t-distribution with \(n_1+n_2-2\) degrees of freedom.
Hypothesis Tests for \(\boldsymbol{\mu_1-\mu_2}\): The Pooled t-test
Now let's consider the hypothesis test for the mean differences with pooled variances.
\(H_0\colon\mu_1-\mu_2=0\)
\(H_a\colon \mu_1-\mu_2\ne0\)
\(H_a\colon \mu_1-\mu_2>0\)
\(H_a\colon \mu_1-\mu_2<0\)
The assumptions/conditions are:
- The populations are independent
- The population variances are equal
- Each population is either normal or the sample size is large.
The test statistic is...
\(t^*=\dfrac{\bar{x}_1-\bar{x}_2-0}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}\)
And \(t^*\) follows a t-distribution with degrees of freedom equal to \(df=n_1+n_2-2\).
The p-value, critical value, rejection region, and conclusion are found similarly to what we have done before.
Example 7-4: Comparing Packing Machines
In a packing plant, a machine packs cartons with jars. It is supposed that a new machine will pack faster on the average than the machine currently used. To test that hypothesis, the times it takes each machine to pack ten cartons are recorded. The results, (machine.txt), in seconds, are shown in the tables.
| 42.1 | 41.3 | 42.4 | 43.2 | 41.8 |
| 41.0 | 41.8 | 42.8 | 42.3 | 42.7 |
\(\bar{x}_1=42.14, \text{s}_1= 0.683\)
| 42.7 | 43.8 | 42.5 | 43.1 | 44.0 |
| 43.6 | 43.3 | 43.5 | 41.7 | 44.1 |
\(\bar{x}_2=43.23, \text{s}_2= 0.750\)
Do the data provide sufficient evidence to conclude that, on the average, the new machine packs faster?
Are these independent samples? Yes, since the samples from the two machines are not related.
Are these large samples or a normal population?
We have \(n_1\lt 30\) and \(n_2\lt 30\). We do not have large enough samples, and thus we need to check the normality assumption from both populations. Let's take a look at the normality plots for this data:
From the normal probability plots, we conclude that both populations may come from normal distributions. Remember the plots do not indicate that they DO come from a normal distribution. It only shows if there are clear violations. We should proceed with caution.
Do the populations have equal variance? No information allows us to assume they are equal. We can use our rule of thumb to see if they are “close.” They are not that different as \(\dfrac{s_1}{s_2}=\dfrac{0.683}{0.750}=0.91\) is quite close to 1. This assumption does not seem to be violated.
We can thus proceed with the pooled t-test.
Let \(\mu_1\) denote the mean for the new machine and \(\mu_2\) denote the mean for the old machine.
The null hypothesis is that there is no difference in the two population means, i.e.
\(H_0\colon \mu_1-\mu_2=0\)
The alternative is that the new machine is faster, i.e.
\(H_a\colon \mu_1-\mu_2<0\)
The significance level is 5%. Since we may assume the population variances are equal, we first have to calculate the pooled standard deviation:
\begin{align} s_p&=\sqrt{\frac{(n_1-1)s^2_1+(n_2-1)s^2_2}{n_1+n_2-2}}\\ &=\sqrt{\frac{(10-1)(0.683)^2+(10-1)(0.750)^2}{10+10-2}}\\ &=\sqrt{\dfrac{9.261}{18}}\\ &=0.7173 \end{align}
The test statistic is:
\begin{align} t^*&=\dfrac{\bar{x}_1-\bar{x}_2-0}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}\\ &=\dfrac{42.14-43.23}{0.7173\sqrt{\frac{1}{10}+\frac{1}{10}}}\\&=-3.398 \end{align}
The alternative is left-tailed so the critical value is the value \(a\) such that \(P(T<a)=0.05\), with \(10+10-2=18\) degrees of freedom. The critical value is -1.7341. The rejection region is \(t^*<-1.7341\).
Our test statistic, -3.3978, is in our rejection region, therefore, we reject the null hypothesis. With a significance level of 5%, we reject the null hypothesis and conclude there is enough evidence to suggest that the new machine is faster than the old machine.
To find the interval, we need all of the pieces. We calculated all but one when we conducted the hypothesis test. We only need the multiplier. For a 99% confidence interval, the multiplier is \(t_{0.01/2}\) with degrees of freedom equal to 18. This value is 2.878.
The interval is:
\(\bar{x}_1-\bar{x}_2\pm t_{\alpha/2}s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}\)
\((42.14-43.23)\pm 2.878(0.7173)\sqrt{\frac{1}{10}+\frac{1}{10}}\)
\(-1.09\pm 0.9232\)
The 99% confidence interval is (-2.013, -0.167).
We are 99% confident that the difference between the two population mean times is between -2.012 and -0.167.
Minitab: 2-Sample t-test - Pooled
The following steps are used to conduct a 2-sample t-test for pooled variances in Minitab.
- Choose Stat > Basic Statistics > 2-Sample t .
- The following dialog boxes will then be displayed.
Note! When entering values into the samples in different columns input boxes, Minitab always subtracts the second value (column entered second) from the first value (column entered first). - Select the Options button and enter the desired 'confidence level', 'null hypothesis value' (again for our class this will be 0), and select the correct 'alternative hypothesis' from the drop-down menu. Finally, check the box for 'assume equal variances'. This latter selection should only be done when we have verified the two variances can be assumed equal.
The Minitab output for the packing time example:
Two-Sample T-Test and CI: New Machine, Old Machine
Method
μ1: mean of New Machine
μ2: mean of Old Machine
Difference: μ1 - μ2
Equal variances are assumed for this analysis.
Descriptive Statistics
| Sample |
N |
Mean |
StDev |
SE Mean |
|
New Machine |
10 |
42.140 |
0.683 | 0.22 |
|
Old Machine |
10 |
43.230 |
0.750 | 0.24 |
Estimation for Difference
| Difference | Pooled StDev | 95% Upper Bound for Difference |
|
-1.090 |
0.717 | -0.534 |
Test
Alternative hypothesis
H1: μ1 - μ2 < 0
| T-Value | DF | P-Value |
|---|---|---|
| -3.40 | 18 | 0.002 |
7.3.1.2 - Unpooled Variances
7.3.1.2 - Unpooled VariancesWhen the assumption of equal variances is not valid, we need to use separate, or unpooled, variances. The mathematics and theory are complicated for this case and we intentionally leave out the details.
We still have the following assumptions:
- The populations are independent
- Each population is either normal or the sample size is large.
If the assumptions are satisfied, then
\(t^*=\dfrac{\bar{x}_1-\bar{x_2}-0}{\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}}\)
will have a t-distribution with degrees of freedom
\(df=\dfrac{(n_1-1)(n_2-1)}{(n_2-1)C^2+(1-C)^2(n_1-1)}\)
where \(C=\dfrac{\frac{s^2_1}{n_1}}{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}\).
- \((1-\alpha)100\%\) Confidence Interval for \(\mu_1-\mu_2\) for Unpooled Variances
- \(\bar{x}_1-\bar{x}_2\pm t_{\alpha/2} \sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}\)
-
Where \(t_{\alpha/2}\) comes from the t-distribution using the degrees of freedom above.
Minitab®
Minitab: Unpooled t-test
To perform a separate variance 2-sample, t-procedure use the same commands as for the pooled procedure EXCEPT we do NOT check box for 'Use Equal Variances.'
- Choose Stat > Basic Statistics > 2-sample t
- Select the Options box and enter the desired 'Confidence level,' 'Null hypothesis value' (again for our class this will be 0), and select the correct 'Alternative hypothesis' from the drop-down menu.
- Choose OK.
For some examples, one can use both the pooled t-procedure and the separate variances (non-pooled) t-procedure and obtain results that are close to each other. However, when the sample standard deviations are very different from each other, and the sample sizes are different, the separate variances 2-sample t-procedure is more reliable.
Example 7-5: Grade Point Average
Independent random samples of 17 sophomores and 13 juniors attending a large university yield the following data on grade point averages (student_gpa.txt):
| 3.04 | 2.92 | 2.86 | 1.71 | 3.60 |
| 3.49 | 3.30 | 2.28 | 3.11 | 2.88 |
| 2.82 | 2.13 | 2.11 | 3.03 | 3.27 |
| 2.60 | 3.13 |
| 2.56 | 3.47 | 2.65 | 2.77 | 3.26 |
| 3.00 | 2.70 | 3.20 | 3.39 | 3.00 |
| 3.19 | 2.58 | 2.98 |
At the 5% significance level, do the data provide sufficient evidence to conclude that the mean GPAs of sophomores and juniors at the university differ?
There is no indication that there is a violation of the normal assumption for both samples. As before, we should proceed with caution.
Now, we need to determine whether to use the pooled t-test or the non-pooled (separate variances) t-test. The summary statistics are:
|
Variable |
Sample size |
Mean |
Standard Deviation |
|---|---|---|---|
| sophomore |
17 |
2.840 |
0.52 |
|
junior |
13 |
2.981 |
0.3093 |
The standard deviations are 0.520 and 0.3093 respectively; both the sample sizes are small, and the standard deviations are quite different from each other. We, therefore, decide to use an unpooled t-test.
The null and alternative hypotheses are:
\(H_0\colon \mu_1-\mu_2=0\) vs \(H_a\colon \mu_1-\mu_2\ne0\)
The significance level is 5%. Perform the 2-sample t-test in Minitab with the appropriate alternative hypothesis.
Remember, the default for the 2-sample t-test in Minitab is the non-pooled one. Minitab generates the following output.
Two sample T for sophomores vs juniors
| N | Mean | StDev | SE Mean | |
|---|---|---|---|---|
| sophomore | 17 | 2.840 | 0.52 | 0.13 |
| junior | 13 | 2.981 | 0.309 | 0.086 |
95% CI for mu sophomore - mu juniors: (-0.45, 0.173)
T-Test mu sophomore = mu juniors (Vs no =): T = -0.92
P = 0.36 DF = 26
Since the p-value of 0.36 is larger than \(\alpha=0.05\), we fail to reject the null hypothesis.
At 5% level of significance, the data does not provide sufficient evidence that the mean GPAs of sophomores and juniors at the university are different.
95% CI for mu sophomore- mu juniors is;
(-0.45, 0.173)
We are 95% confident that the difference between the mean GPA of sophomores and juniors is between -0.45 and 0.173.
7.3.2 - Inference for Paired Means
7.3.2 - Inference for Paired MeansIntroduction
When we developed the inference for the independent samples, we depended on the statistical theory to help us. The theory, however, required the samples to be independent. What can we do when the two samples are not independent, i.e., the data is paired?
Consider an example where we are interested in a person’s weight before implementing a diet plan and after. Since the interest is focusing on the difference, it makes sense to “condense” these two measurements into one and consider the difference between the two measurements. For example, if instead of considering the two measures, we take the before diet weight and subtract the after diet weight. The difference makes sense too! It is the weight lost on the diet.
When we take the two measurements to make one measurement (i.e., the difference), we are now back to the one sample case! Now we can apply all we learned for the one sample mean to the difference (Cool!)
The Confidence Interval for the Difference of Paired Means, \(\mu_d\)
When we consider the difference of two measurements, the parameter of interest is the mean difference, denoted \(\mu_d\). The mean difference is the mean of the differences. We are still interested in comparing this difference to zero.
Suppose we have two paired samples of size \(n\):
\(x_1, x_2, …., x_n\) and \(y_1, y_2, … , y_n\)
Their difference can be denoted as:
\(d_1=x_1-y_1, d_2=x_2-y_2, …., d_n=x_n-y_n\)
The sample mean of the differences is:
\(\bar{d}=\frac{1}{n}\sum_{i=1}^n d_i\)
Denote the sample standard deviation of the differences as \(s_d\).
If \(\bar{d}\) is normal (or the sample size is large), the sampling distribution of \(\bar{d}\) is (approximately) normal with mean \(\mu_d\), standard error \(\dfrac{\sigma_d}{\sqrt{n}}\), and estimated standard error \(\dfrac{s_d}{\sqrt{n}}\).
At this point, the confidence interval will be the same as that of one sample.
- \(\boldsymbol{(1-\alpha)100\%}\) Confidence interval for \(\boldsymbol{\mu_d}\)
-
\(\bar{d}\pm t_{\alpha/2}\frac{s_d}{\sqrt{n}}\)
where \(t_{\alpha/2}\) comes from \(t\)-distribution with \(n-1\) degrees of freedom
Example 7-6: Zinc Concentrations
Trace metals in drinking water affect the flavor and an unusually high concentration can pose a health hazard. Ten pairs of data were taken measuring zinc concentration in bottom water and surface water (zinc_conc.txt).
Does the data suggest that the true average concentration in the bottom water is different than that of surface water? Construct a confidence interval to address this question.
Zinc concentrations
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | |
|---|---|---|---|---|---|---|---|---|---|---|
| Bottom Water | .430 | .266 | .567 | .531 | .707 | .716 | .651 | .589 | .469 | .723 |
| Surface Water | .415 | .238 | .390 | .410 | .605 | .609 | .632 | .523 | .411 | .612 |
In this example, the response variable is concentration and is a quantitative measurement. The explanatory variable is location (bottom or surface) and is categorical. The two populations (bottom or surface) are not independent. Therefore, we are in the paired data setting. The parameter of interest is \(\mu_d\).
Find the difference as the concentration of the bottom water minus the concentration of the surface water.
Since the problem did not provide a confidence level, we should use 5%.
To use the methods we developed previously, we need to check the conditions. The problem does not indicate that the differences come from a normal distribution and the sample size is small (n=10). We should check, using the Normal Probability Plot to see if there is any violation. First, we need to find the differences.
|
Difference |
0.015 |
0.028 |
0.177 |
0.121 |
0.102 |
0.107 |
0.019 |
0.066 |
0.058 |
0.111 |
|---|
All of the differences fall within the boundaries, so there is no clear violation of the assumption. We can proceed with using our tools, but we should proceed with caution.
We need all of the pieces for the confidence interval. The sample mean difference is \(\bar{d}=0.0804\) and the standard deviation is \(s_d=0.0523\). For practice, you should find the sample mean of the differences and the standard deviation by hand. With \(n-1=10-1=9\) degrees of freedom, \(t_{0.05/2}=2.2622\).
The 95% confidence interval for the mean difference, \(\mu_d\) is:
\(\bar{d}\pm t_{\alpha/2}\dfrac{s_d}{\sqrt{n}}\)
\(0.0804\pm 2.2622\left( \dfrac{0.0523}{\sqrt{10}}\right)\)
(0.04299, 0.11781)
We are 95% confident that the population mean difference of bottom water and surface water zinc concentration is between 0.04299 and 0.11781.
If there is no difference between the means of the two measures, then the mean difference will be 0. Since 0 is not in our confidence interval, then the means are statistically different (or statistical significant or statistically different).
Note! Minitab will calculate the confidence interval and a hypothesis test simultaneously. We demonstrate how to find this interval using Minitab after presenting the hypothesis test.
Hypothesis Test for the Difference of Paired Means, \(\mu_d\)
In this section, we will develop the hypothesis test for the mean difference for paired samples. As we learned in the previous section, if we consider the difference rather than the two samples, then we are back in the one-sample mean scenario.
The possible null and alternative hypotheses are:
\(H_0\colon \mu_d=0\)
\(H_a\colon \mu_d\ne 0\)
\(H_a\colon \mu_d>0\)
\(H_a\colon \mu_d<0\)
We still need to check the conditions and at least one of the following need to be satisfied:
- The differences of the paired follow a normal distribution
- The sample size is large, \(n>30\).
If at least one is satisfied then...
\(t^*=\dfrac{\bar{d}-0}{\frac{s_d}{\sqrt{n}}}\)
Will follow a t-distribution with \(n-1\) degrees of freedom.
The same process for the hypothesis test for one mean can be applied. The test for the mean difference may be referred to as the paired t-test or the test for paired means.
Example 7-7: Zinc Concentrations - Hypothesis Test
Recall the zinc concentration example. Does the data suggest that the true average concentration in the bottom water exceeds that of surface water? Conduct this test using the rejection region approach. (zinc_conc.txt).
Zinc concentrations
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | |
|---|---|---|---|---|---|---|---|---|---|---|
| Bottom Water | .430 | .266 | .567 | .531 | .707 | .716 | .651 | .589 | .469 | .723 |
| Surface Water | .415 | .238 | .390 | .410 | .605 | .609 | .632 | .523 | .411 | .612 |
If we find the difference as the concentration of the bottom water minus the concentration of the surface water, then null and alternative hypotheses are:
\(H_0\colon \mu_d=0\) vs \(H_a\colon \mu_d>0\)
Note! If the difference was defined as surface - bottom, then the alternative would be left-tailed.
The desired significance level was not stated so we will use \(\alpha=0.05\).
The assumptions were discussed when we constructed the confidence interval for this example. Remember although the Normal Probability Plot for the differences showed no violation, we should still proceed with caution.
The next step is to find the critical value and the rejection region. The critical value is the value \(a\) such that \(P(T>a)=0.05\). Using the table or software, the value is 1.8331. For a right-tailed test, the rejection region is \(t^*>1.8331\).
Recall from the previous example, the sample mean difference is \(\bar{d}=0.0804\) and the sample standard deviation of the difference is \(s_d=0.0523\). Therefore, the test statistic is:
\(t^*=\dfrac{\bar{d}-0}{\frac{s_d}{\sqrt{n}}}=\dfrac{0.0804}{\frac{0.0523}{\sqrt{10}}}=4.86\)
The value of our test statistic falls in the rejection region. Therefore, we reject the null hypothesis. With a significance level of 5%, there is enough evidence in the data to suggest that the bottom water has higher concentrations of zinc than the surface level.
Minitab® – Paired t-Test
You can use a paired t-test in Minitab to perform the test. Alternatively, you can perform a 1-sample t-test on difference = bottom - surface.
- Choose Stat> Basic Statistics> Paired t
- Click Optionsto specify the confidence level for the interval and the alternative hypothesis you want to test. The default null hypothesis is 0.
Zinc Concentrations Example
The Minitab output for paired T for bottom - surface is as follows:
Paired T for bottom - surface
|
N |
Mean |
StDev |
SE Mean |
|
|
bottom |
10 |
0.5649 |
0.1468 |
0.0464 |
|
surface |
10 |
0.4845 |
0.1312 |
0.0415 |
|
Difference |
10 |
0.0804 |
0.0523 |
0.0165 |
95% lower bound for mean difference: 0.0505
T-Test of mean difference = 0 (vs > 0): T-Value = 4.86 P-Value = 0.000
Note! In Minitab, if you choose a lower-tailed or an upper-tailed hypothesis test, an upper or lower confidence bound will be constructed, respectively, rather than a confidence interval.
Using the p-value to draw a conclusion about our example:
p-value = 0.000 < 0.05
Reject \(H_0\) and conclude that bottom zinc concentration is higher than surface zinc concentration.
Additional Notes
- For the zinc concentration problem, if you do not recognize the paired structure, but mistakenly use the 2-sample t-test treating them as independent samples, you will not be able to reject the null hypothesis. This demonstrates the importance of distinguishing the two types of samples. Also, it is wise to design an experiment efficiently whenever possible.
- What if the assumption of normality is not satisfied? Considering a nonparametric test would be wise.