# Suggested answer to exercise: Normal plot for angina and pronethalol, 8

Question 8: Compare the square root transformed numbers of attacks on the two drug treatments using the paired t test. How does the result compare with those from the sign test and from the t test on the raw data?

There are two ways to do this.

1. We can calculate two new variables, square root of number of attacks on placebo and square root of number of attacks on pronethalol, and proceed as in Question 1,
2. We can use the difference we have already calculated and do a one-sample t test of the null hypothesis that the mean difference is zero.

If we were to proceed as we have so far, we would choose option 1. We create two new variables, placeboroot = SQRT(placebo), and pronethroot = SQRT(proneth), using the same method as in Question 2. We then do the paired t test as in Question 1 and get the following output:

Paired Samples Statistics
Mean N Std. Deviation Std. Error Mean
Pair 1 placeboroot 5.8454 12 4.58237 1.32282
pronethroot 4.7032 12 5.07722 1.46567

Paired Samples Correlations
N Correlation Sig.
Pair 1 placeboroot & pronethroot 12 .988 .000

Paired Samples Test
Paired Differences t   df   Sig. (2-tailed)
Mean Std. Deviation Std. Error Mean 95% Confidence Interval of the Difference
Lower Upper
Pair 1 placeboroot - pronethroot 1.14219 .90408 .26099 .56776 1.71661 4.376 11 .001

If we were to use the the second approach, the one sample t test using the differences, we proceed as follows. Click Analyze, Compare Means, One-Sample T Test. Choose variable 'Difference, square root number of attacks'. Click OK.

We should get the following output:

One-Sample Statistics
N Mean Std. Deviation Std. Error Mean
Difference, square root number of attacks 12 1.1422 .26099 .90408 .26099

One-Sample Test
Test Value = 0
t df Sig. (2-tailed) Mean Difference 95% Confidence Interval of the Difference
Upper Lower
Difference, square root number of attacks 4.376 11 .001 1.14219 .5678 1.7166

The mean difference, its standard error, the confidence interval, and the P value are the same either way.

Using the square root transformed data, the P value is 0.001, smaller than but quite similar to the 0.0063 for the sign test. Both are significant. Compare this to the test for the untransformed data, which gave P = 0.107. By comparison, the square root transformed data give a much smaller P value and a different conclusion.

If the assumptions are met and the test is valid, the paired t test is more powerful than the sign test and has a better chance of finding a difference when there really is a difference in the population.

When the assumptions are not met, the t test may be less powerful than less demanding alternative methods.