From these, we obtain the power plot shown in Figure 7. Shortly we will consider some of the ways of increasing power to more acceptable levels.įor each value of μ 1 ≥ 52 we can repeat the above calculations to obtain a value of power. 1136 (cell G9 of Figure 6a) 12.9% of the time, which is quite poor. The specific test that was conducted did not reject the null hypothesis, but we also see that such a test would only have found a very small effect of size. Or T.DIST( t, df, TRUE) in releases of Excel starting with Excel 2010. Note that in Excel 2007 we calculate p-value = TDIST( t, df, 2) = TDIST(3.66, 39, 2) =.
926 2.02 once again we reject the null hypothesis. Note that if we had used the normal distribution for the hypothesis testing as described in Sampling Distributions we would have gotten the following results: This means there is an 8.8% probability of achieving a value for t this high assuming that the null hypothesis is true, and since 8.8% > 5% we can’t reject the null hypothesis. 05 = α, the null hypothesis is not rejected. Since Excel only displays the values of these formulas, we show each of the formulas (in text format) in column G so that you can see how the calculations are performed. the program is not effectiveįrom the box plot in Figure 2 we see that the data is quite symmetric and so we use the t-test even though the sample is small.Ĭolumn E of Figure 1 contains all the formulas required to carry out the t test. from an earlier study, that overall weight gain is unlikely). Usually, we conduct a two-tailed test since there is a risk that the program might actual result in weight gain rather than loss, but for this example, we will conduct a one-tailed test (perhaps because we have evidence, e.g. We judge the program to be effective if there is some weight loss at the 95% significance level. Can we conclude that the program is effective?Ī negative value in column B indicates that the subject gained weight. To test this claim 12 people were put on the program and the number of pounds of weight gain/loss was recorded for each person after two years, as shown in columns A and B of Figure 1.
The mean is approximately equal to the median.the median is in the center of the box and the whiskers extend equally in each direction
The boxplot is relatively symmetrical i.e.The following are indications of symmetry: This can be determined by graphing the data. It turns out that the t distribution provides good results even when the population is not normal and even when the sample is small, provided the sample data is reasonably symmetrically distributed about the sample mean. The t distribution provides a good way to perform one-sample tests on the mean when the population variance is not known provided the population is normal or the sample is sufficiently large so that the Central Limit Theorem applies (see Theorem 1 and Corollary 1 of Basic Concepts of t Distribution).