tests involving paired differences

Question

You wish to test the following claim (HaHa) at a significance level of α=0.01. For the context of this problem, μd=μ2-μ1 where the first data set represents a pre-test and the second data set represents a post-test.

      Ho:μd=0 

      Ha:μd≠0

You believe the population of difference scores is normally distributed, but you do not know the standard deviation. You obtain pre-test and post-test samples for n=54 subjects. The average difference (post - pre) is d¯=2.7 with a standard deviation of the differences of sd=28.4.

What is the critical value for this test? (Report answer accurate to three decimal places.) 

critical value = ±±

What is the test statistic for this sample? (Report answer accurate to three decimal places.) 

test statistic =

Answer 1

α = .01

This is a 2-sided test. As n=54, df =n-1 =53

Then, from Excel, t.inv.2t(.01,53) = 2.672.

The test statistic t= d/(σ/√n) = 2.7 √54/28.4 = .699

As t < t_crit, we fail to reject the null hypothesis of no difference.

Answer 2

Hi Shelby!

A paired difference t-test is tested in the same way as a one-sample t-test.

Critical Value:

Critical values allow us to compare how far away our sample is from our expected distribution. If our test statistic larger than our critical value, then we would think our sample is really different from what we expect. If it is smaller than our critical value, then our sample may be representative of the population.

Our critical value can be identified by our significance level (α = 0.01) and our degrees of freedom (df = n-1). Since our alternative hypothesis is μd ≠ 0, we know we are conducting a two-sided t-test.

This means we will want to look up the value at 0.995 with 53 degrees of freedom.

1. Because we have a two-sided test, we want the total area under each tail to equal our alpha. So when we look at a one sided table, we have 1 - (0.01 / 2) = 0.995.
2. Our degrees of freedom is 54 - 1 = 53.

Now that you've done the hard part, we can do the easy part which is to find the value! You can look up the value in your stats textbook or online (https://goodcalculators.com/student-t-value-calculator/) .

Your critical value should be ± 2.672.

Test-Statistic:

In order to calculate our test statistic or t-score, we can use the following formula:

t-statistics = (sample diff - true diff) / standard error

1. Numerator: The average difference from our sample is 2.7. We are testing against the null hypothesis that the true difference is 0. Our numerator is (2.7 - 0).
2. Denominator: standard error = sd / √n. Our standard deviation = 28.4 and our sample size is 54. Our denominator is (28.4 / √54).

As a result our t-statistic is 2.7 / (28.4 / √54) = 0.699

One step further... :)

While your problem doesn't ask this specifically, it is still helpful to understand how we should interpret this!

We can compare our t-statistic with our critical value to tell us whether or not we should accept or reject our null hypothesis.

Here, we find that our t-statistic is smaller than our critical value. (0.699 < 2.672)

As a result, we do not have sufficient evidence to reject our null hypothesis (it is likely the students did not do significantly better or worse on their post-test than their pre-test.). We would fail to reject our null hypothesis in this case.

Good luck!