9.4 Statistical Inference for Matched Samples

In a clinical trial for a new drug, patients are tested before the drug is administered and then the same group of patients are tested after being given the drug.  This is a matched sample experiment because the same group of patients is measured before and after the administration of the drug.  In this way, there are a pair of observations (a before measurement and an after measurement) for each patient.

A manufacturing company wants to know which of two different production methods allow employees to perform a task the fastest.  The table below illustrates the difference in an independent sample design and a matched sample design to test the difference in the average time it takes to perform the task using the two different methods.

In the independent sample design, there is no relationship between the two groups of employees.  In the matched sample design, there is one group of employees with a pair of observations (a time from Method 1 and a time from Method 2) for each employee.

A company has two different methods that employees can use to complete a manufacturing task.  A sample of workers is taken and the time, in minutes, that each worker takes to complete the task using each method is recorded.  The data is shown in the table below.  Assume the differences in the paired times have a normal distribution.

1. Construct a 98% confidence interval for the mean difference in the time it takes the workers to complete the task.
2. Interpret the confidence interval found in part 1.
3. Is there evidence to suggest that the mean completion time for the two methods is the same?  Explain.

Solution:

1. We start by calculating out the differences in the paired data.  We will calculate the differences as Method 1-Method 2.
   

   Worker	Method 1	Method 2	Difference
   1	5.5	6.8	-1.3
   2	6.9	6.6	0.3
   3	6.1	5.1	1
   4	6	6.8	-0.8
   5	7	6.7	0.3
   6	6.7	6.5	0.2
   7	6.4	5.8	0.6
   8	7	6.8	0.2
   9	6.6	5.3	1.3
   10	5.7	5.8	-0.1
   11	5.9	6.9	-1
   12	7	6.7	0.3
   13	5.4	6.5	-1.1
   14	5.4	6.3	-0.9
   15	5.3	5	0.3

   From the difference column, we have [latex]n_D=15[/latex], [latex]\overline{x}_D=-0.0466...[/latex], and [latex]s_D=0.7936...[/latex].

   To find the confidence interval, we need to find the [latex]t[/latex]-score for the 98% confidence interval.  This means that we need to find the [latex]t[/latex]-score so that the sum of the area in the tails is [latex]1-0.98=0.02[/latex].  The degrees of freedom for the [latex]t[/latex]-distribution is [latex]df=n_D-1=15-1=14[/latex].

   

   Function	t.inv.2t	Answer
   Field 1	0.02	2.6244…
   Field 2	14

   So [latex]t=2.6244...[/latex]. The 98% confidence interval is

   [latex]\begin{eqnarray*} \\ \mbox{Lower Limit} & = & \overline{x}_D-t \times \frac{s_D}{\sqrt{n_D}}\\ & = & -0.0466...-2.6244... \times \frac{0.7936...}{\sqrt{15}} \\ & = & -0.584 \\ \\ \mbox{Upper Limit} & = & \overline{x}_D+t \times \frac{s_D}{\sqrt{n_D}}\\ & = & -0.0466...+2.6244... \times \frac{0.7936...}{\sqrt{15}} \\ & = & 0.491  \\ \\ \end{eqnarray*}[/latex]

2. We are 98% confident that the mean difference in the completion times using the two methods is between -0.584 minutes and 0.491 minutes.
3. Because 0 is inside the confidence interval, it suggests that the mean difference [latex]\displaystyle{\mu_D}[/latex] is 0.  That is, [latex]\displaystyle{\mu_D= 0}[/latex].  This suggests that the mean completion times for the two methods are the same.

A study was conducted to investigate the effectiveness of hypnosis on reducing pain.  Eight subjects are randomly selected.  Each subject’s pain is measured before and after being hypnotized.  A lower score indicates less pain. Assume the differences in the before and after scores have a normal distribution.  At the 5% significance level, are the pain sensory measurements, on average, lower after hypnotism?

Solution:

We start by calculating out the differences in the paired data.  We will calculate the differences as before-after.

From the difference column, we have [latex]n_D=8[/latex], [latex]\overline{x}_D=3.125[/latex], and [latex]s_D=2.911...[/latex].

Hypotheses:

[latex]\begin{eqnarray*} H_0: & & \mu_D=0  \\ H_a: & & \mu_D \gt 0  \end{eqnarray*}[/latex]

p-value: 

This is a test on the mean difference in matched samples, so we use a [latex]t[/latex]-distribution to calculate the p-value.  Because the alternative hypothesis is a [latex]\gt[/latex], the p-value is the area in the right tail of the distribution.

To use the t.dist.rt function, we need to calculate out the [latex]t[/latex]-score:

[latex]\begin{eqnarray*} t & = & \frac{\overline{x}_D-\mu_D}{\frac{s_D}{\sqrt{n}_D}} \\ & = & \frac{3.125-0}{\frac{2.911...}{\sqrt{8}}} \\ & = & 3.0359... \end{eqnarray*}[/latex]

The degrees of freedom for the [latex]t[/latex]-distribution is [latex]n_D-1=8-1=7[/latex].

So the p-value[latex]=0.0095[/latex].

Conclusion:

Because p-value[latex]=0.0095 \lt 0.05=\alpha[/latex], we reject the null hypothesis in favour of the alternative hypothesis.  At the 5% significance level there is enough evidence to suggest that, on average, the pain sensory measurements are lower after hypnotism.

A study was conducted to investigate how effective a new diet was in lowering cholesterol. Nine patients were selected for the new diet and their cholesterol was measured before and after starting the new diet.  The results are recorded in the table below.  Assume the differences have a normal distribution.  At the 5% significance level, was the new diet, on average, successful in lowering patients’ cholesterol?

Solution:

We start by calculating out the differences in the paired data.  We will calculate the differences as after-before.

From the difference column, we have [latex]n_D=9[/latex], [latex]\overline{x}_D=-10.777...[/latex], and [latex]s_D=11.861...[/latex].

Hypotheses:

[latex]\begin{eqnarray*} H_0: & & \mu_D=0  \\ H_a: & & \mu_D \lt 0  \end{eqnarray*}[/latex]

p-value: 

This is a test on the mean difference in matched samples, so we use a [latex]t[/latex]-distribution to calculate the p-value.  Because the alternative hypothesis is a [latex]\lt[/latex], the p-value is the area in the left tail of the distribution.

To use the t.dist function, we need to calculate out the [latex]t[/latex]-score:

[latex]\begin{eqnarray*} t & = & \frac{\overline{x}_D-\mu_D}{\frac{s_D}{\sqrt{n}_D}} \\ & = & \frac{-10.777...-0}{\frac{11.861...}{\sqrt{9}}} \\ & = & -2.725... \end{eqnarray*}[/latex]

The degrees of freedom for the [latex]t[/latex]-distribution is [latex]n_D-1=9-1=8[/latex].

So the p-value[latex]=0.0130[/latex].

Conclusion:

Because p-value[latex]=0.0130 \lt 0.05=\alpha[/latex], we reject the null hypothesis in favour of the alternative hypothesis.  At the 5% significance level there is enough evidence to suggest that, on average, the new diet lowered the patients’ cholesterol levels.

Seven eighth graders at Kennedy Middle School measured how far they could push the shot-put with their dominant (writing) hand and their weaker (non-writing) hand.  They thought that they could push equal distances with either hand.  The results from their throws are recorded in the table below.  Assume the differences are normally distributed.  At the 5% significance level, is there a difference in the average distance for the dominant versus weaker hand?

Solution:

We start by calculating out the differences in the paired data.  We will calculate the differences as dominant-weaker.

From the difference column, we have [latex]n_D=7[/latex], [latex]\overline{x}_D=3.714...[/latex], and [latex]s_D=4.498...[/latex].

Hypotheses:

[latex]\begin{eqnarray*} H_0: & & \mu_D=0  \\ H_a: & & \mu_D \neq 0  \end{eqnarray*}[/latex]

p-value: 

This is a test on the mean difference in matched samples, so we use a [latex]t[/latex]-distribution to calculate the p-value.  Because the alternative hypothesis is a [latex]\neq[/latex], the p-value is the sum of area in the tails of the distribution.

To use the t.dist.2t function, we need to calculate out the [latex]t[/latex]-score:

[latex]\begin{eqnarray*} t & = & \frac{\overline{x}_D-\mu_D}{\frac{s_D}{\sqrt{n}_D}} \\ & = & \frac{3.714...-0}{\frac{4.498...}{\sqrt{7}}} \\ & = & 2.184... \end{eqnarray*}[/latex]

The degrees of freedom for the [latex]t[/latex]-distribution is [latex]n_D-1=7-1=6[/latex].

So the p-value[latex]=0.0716[/latex].

Conclusion:

Because p-value[latex]=0.0716 \gt 0.05=\alpha[/latex], we do not reject the null hypothesis.  At the 5% significance level there is not enough evidence to suggest that there a difference in the average distance for the dominant versus weaker hand.