Question

You are a marketing expert for a company that is producing a new kind of energy bar. You want to provide evidence that consuming the new type of bar 1 hour before engaging in strenuous exercise of long duration such as running a marathon-will have an effect on performance. You decide to conduct a test at a half-marathon race. You give the bar to a randomly selected group of 40 partidpants (the treatment group). You randomly select a second group of 40 partidpants, to whom you do not give the energy bar (the control group). The average time for the control group was 2 hours and 20 minutes (2:20, 140 minutes) with a standard deviation of 27 minutes. The average time for the treatment group was 2 hours and 7 minutes (2:07, or 127 minutes) with a standard deviation of 22 minutes. You use a statistical computing package to conduct the t test for independent samples. The following tables represent your output: Group Statistics Sed Frro viation Mean Time Group Control Treatment N Mean 40 140 40127 4.27 2 2 Independent Samples Test Test for Equality of Variances t Test for Equality of Means Sig Mean (2-tailed) Difference 13.00 Std Error Difference sig Time 1.263 363 0.289 3.339 0.001 Equal variances assumed Equal variances not assumed 2.361 0.021 13.00 Looking at the significance (Sig) of the Test for Equality of Variances, what should your action be?

Answer 1

Test for equality of Variance: p-value > 0.05

• At α = 0.05, this result is non-significant, meaning that the data does not provide evidence that the variances are different. Thus you can assume equal variance and use the "Equal variance assumed" row of the table.

Most accurate statement for marketing:

• Consuming the energy bar 1 hour before the race had a significant effect on performance (t(78) = 3.339), p < 0.05). Participants ran an average of thirteen minutes faster when they consumed the energy bar.