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Wize University Statistics Textbook > Match Pairs (Dependent Means)

Hypothesis Testing for Matched Pairs

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Hypothesis Test for Matched Pairs

When a hypothesis test is performed for matched pairs, the subjects or observations are matched in pairs and then we analyze the differences.


Wize Tip
Review Hypothesis Testing if you need a refresher of the five steps. (See: Hypothesis Testing with One Sample)


Hypotheses:

Ho:D‾=0H_o:\overline{D}=0 (there is no difference in the population mean of the differences)
Ha:D‾≠0H_a:\overline{D}\ne0 (there is a difference in the population mean of the differences)


Wize Concept
Some textbooks use the following notations:

Ho:μd=0H_o:\mu_d=0 (there is no difference in the population mean of the differences)
Ha:μd≠0H_a:\mu_d\ne0 (there is a difference in the population mean of the differences)


Basically, we compute the difference between the two means and plug the value into a one-sample tt-test on the differences.
  • If the mean difference is zero, then the two means are not different.
  • If the mean difference is statistically significantly not zero, then the two means are different.

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d‾\overline{d} is the sample mean of the differences
d‾=x‾1i−x‾2i=∑din\overline{d}=\overline{x}_{1i}-\overline{x}_{2i}=\frac{\sum_{ }^{ }d_i}{n}
sds_d is the sample standard deviation of the differences
sd=∑(di−d‾)2n−1s_d=\sqrt{\frac{\sum_{ }^{ }\left(d_i-\overline{d}\right)^2}{n-1}}

Test-statistic:
t=d‾−Dˉsdn\displaystyle\boxed{t=\frac{\overline{d}-\bar D}{\frac{s_d}{\sqrt{n}}}}
where Dˉ=0\bar D=0 such that:

t=d‾−0sdn\displaystyle\boxed{t=\frac{\overline{d}-0}{\frac{s_d}{\sqrt{n}}}}

Degrees of freedom:
n−1n-1
where nn = number of pairs (or sample size)


Wize Concept
Some textbook uses the following notations:
t=x‾d−μdsdn\displaystyle\boxed{t=\frac{\overline{x}_d-\mu_d}{\frac{s_d}{\sqrt{n}}}}
where μd=0\mu_d=0 such that:
t=x‾d−0sdn\displaystyle\boxed{t=\frac{\overline{x}_d-0}{\frac{s_d}{\sqrt{n}}}}


Wize Tip
How can you tell if two mean are dependent of each other, thereby requiring you to do a matched pairs tt-test? A classic example is when you are testing the same person before and after a treatment.

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Example
Hank's Auto Mall sells used cars and got them all appraised by Auto Corp. He hired another car appraisal company, Vroom Car Appraisal, to get a second opinion on those same cars. Hank wants to see if there is a significant difference between the appraisal values for each of his cars. Is there evidence at the 5% significance level? Eight cars were randomly sampled. Values are in $,000's:

Auto CorpVroomDifferencex1x2di=x1−x2CAR 15158−7CAR 275669CAR 3876225CAR 4705713CAR 54951−2CAR 6827111CAR 7604218CAR 8604713Meanx‾1=66.75x‾2=56.75d‾=10Standard Deviations1=14.02s2=9.74sd=10.3\begin{array}{|l|c|c|c|}\hline & \text{Auto Corp} & \text{Vroom} & \text{Difference}\\ &x_1 & x_2 & d_i=x_1-x_2\\\hline \text{CAR 1} & 51 & 58 & -7\\\hline \text{CAR 2} & 75 & 66 & 9\\\hline \text{CAR 3} & 87 & 62 & 25\\\hline \text{CAR 4} & 70 & 57 & 13\\\hline \text{CAR 5} & 49 & 51 & -2\\\hline \text{CAR 6} & 82 & 71 & 11\\\hline \text{CAR 7} & 60 & 42 & 18\\\hline \text{CAR 8} & 60 & 47 & 13\\\hline \text{Mean} & \overline{x}_1=66.75 & \overline{x}_2=56.75 & \overline{d}=10\\\hline \text{Standard Deviation} & s_1=14.02 & s_2=9.74 & s_d=10.3\\\hline \end{array}

The mean appraisal values for each company are dependent on each other because they are based on the same 8 cars - each car was measured twice. In other words, the cars sampled were not drawn from different, independent populations. Rather, they are drawn from the sample population.

Sample mean of the differences, d‾=10\overline{d}=10
Sample standard deviation of the differences, sd=10.3s_d= 10.3
Sample size, n=8n=8 (not 16!)

What does d‾\overline{d} mean in this example (in plain English)?
The mean difference in appraisal value between the two appraisers is $10,000.

Hypotheses:
Ho:Dˉ=0H_o:\bar D=0
Ha:Dˉ≠0H_a:\bar D\ne0 (two-tail test)


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Test-statistic:
t=d‾−0sdn=1010.3/8=2.75t=\dfrac{\overline{d}-0}{\dfrac{s_d}{\sqrt{n}}}=\dfrac{10}{10.3/\sqrt{8}}=2.75
df=n−1=8−1=7df=n-1=8-1=7

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P-value:



t-table: 0.02<0.02< p-value <0.05<0.05
Software: p-value = 0.0287

Since the pp-value is less than α=0.05\alpha=0.05, we have evidence that there is a difference between the two appraisal values between the two appraisers.

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What happens if we incorrectly do a two-sample test, given the paired nature of the data?


Since the variances are quite equal, suppose we incorrectly do a pooled two-sample test for independent means:

sp=12.07s_p=12.07
n1=8n_1=8
n2=8n_2=8
(Notice that now it looks like there are 16 cars being appraised, 8 each from 2 different populations!)


t=(x‾1–x‾2)−0sp1n1+1n2=(66.75–56.75)12.0718+18=1.66t =\dfrac{(\overline{x}_1 – \overline{x}_2 )-0}{s_p \sqrt{\dfrac{1}{n_1} +\dfrac{1}{n_2} }}=\dfrac{(66.75– 56.75)}{12.07\sqrt{\dfrac{1}{8}+\dfrac{1}{8}}}=1.66


df=8+8–2=14df = 8 + 8 – 2 = 14
Using software: pp-value = 0.1198

In this case, the pp-value is greater than α=0.05\alpha=0.05. Doing a two-sample test (incorrectly) suggests that no evidence that the two means differ.


Watch Out!
It is important that you can tell the difference between dependent means or independent means.

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Example: Hypothesis Test for Matched Pairs


The cabbage diet, invented by Dr. Patch, aims at helping people lose weight. According to this diet, eating lots of cabbage soup can reduce your consumption of calories. Six people are asked to track the number of calories consumed during a regular week and a week when they were on the cabbage diet. The number of calories consumed is recorded below:


(a) State the null and alternative hypotheses.

Di=D_i=Regular week −- Cabbage Diet Week

Ho:Dˉ=0    Ha:Dˉ>0H_o:\bar D=0\space\space\space\space H_a:\bar D>0 (one-tail test)

or

Ho:μd=0    Ha:μd>0H_o:\mu_d=0\space\space\space\space H_a:\mu_d>0 (one-tail test)


(b) Compute the test-statistic

t⋅=178.33−0193.436=2.258\displaystyle{t^{\cdot}=\frac{178.33-0}{\frac{193.43}{\sqrt{6}}}=2.258}

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(c) What the p-value?

df=6−1=5df = 6 -1 = 5



2.015<t<2.5712.015 < t < 2.571 so the one-sided p-value is between 0.025 and 0.05.
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(d) At the 5% significance level, is there evidence that this diet reduces the consumption of calories?

Since the p-value is between 2.5% and 5%, it is less than 5% so reject HoH_o.
Thus, there is evidence that this diet works.


(e) At the 1% significance level, is there evidence that this diet reduces the consumption of calories?

Since the p-value is between 2.5% and 5%, it is greater than 1% so we fail to reject HoH_o.
Thus, there is no strong evidence that this diet works.


Twin Pines Winery takes their visitors to the gift shop first before the wine tasting. After the wine tasting, they bring them back to the gift shop. We want to compare the difference in spending at the gift shop before vs. after visitors experience the wine tasting. On average, visitors spend $1.78 more at the gift shop after the wine tasting with a standard deviation of $7.73. This is based on a random sample of 51 visitors.
(i) What is the critical value (t⋅t^{\cdot}) if the significance level is α=0.05\alpha=0.05?
Extra Practice
Hypothesis Testing: Matched Pairs
Seven hockey players try two types of sticks. The number of goals scored using both sticks is recorded for all seven players. Here are the results:
We wish to test if the players will score on average more goals using stick A
Using a p-value of 0.0027, the appropriate conclusion at the 1% level of significance is
Hypothesis Testing: Matched Pairs
In a test, to see if the mean of one population is different than the mean of another population, the appropriate Ho and Ha are
Hypothesis Testing for Matched Pairs
You wish to test if being a member of a union increases hourly wages, in dollars, for the average construction worker. A sample of 18 construction workers' earnings were recorded prior to and after forming a union.
The results are: mean difference = −4.135-4.135 standard deviation of differences = 7.197.19
Carry out a test to see if there is a difference in hourly wages
Matched Pairs for Two Dependent Means
We wonder if judging differs when a judge give scores during the preliminary competition vs. the finale for talent shows. A random sample of 16 judges reveals the following data:
(a) At the 1% significance level, is there is a significant difference between the prices people are willing to pay based on the posters they are shown?
(b) Explain why we are not doing a two-sample test.
Hypothesis testing for Matched Pairs
A matched pairs design hypothesis test is conducted to see if having access to a practice midterm helps improve your test score. The following data is recorded:
A total of 40 students participated in the study.
1. Calculate the test-statistic?
Confidence Intervals for Matched Pairs
Copeman is a private medical clinic. It has a total of 185 locations in B.C. and Ontario. In their efforts to rebrand the company to make it look more luxurious, all the locations were renovated last year. The executives randomly sampled 25 locations from each province and compared their respective sales revenues before and after renovations. Results:
Sample mean of the differences, $5,285
Sample standard deviation of the differences, $25,875
Hypothesis testing for matched pairs
The CEO of ABC Coffee Shop wishes to know if training their employees will make them quicker at serving customers. He randomly selects 7 employees and records the number of hourly customers served without training for four days and with training for another four day period. Here are his results:
a) At the 10% level of significance, what should the CEO conclude?
Hypothesis testing for matched pairs
Copeman is a private medical clinic in B.C. and Ontario. In their efforts to rebrand the company to make it look more luxurious, all the locations were renovated last year. The executives randomly sampled 25 locations from each province and compared their respective sales revenues before and after renovations. Results:
Sample mean of the differences = $5,285
Sample standard deviation of the differences = $25,875