nbld 6061148mbNodeld 2218741998ideploymentid 56870710481 , MINOTAP Complete: Cha

Question

nbld 6061148mbNodeld 2218741998ideploymentid 56870710481 , MINOTAP Complete: Chapter 11 Problem Set Back to Assignment Attempts Average: 12 2. Interpreting statistical software output for an independent-measures t test a Aa 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 randomiy selected group of 10 participants (the treatment group). You randomly select a second group of 10 participants, to whom you do not give the energy bar (the control group). The average time for the control group was 2 hours and 7 minutes (2:07, or 127 minutes) with a standard deviation of 29 minutes. The average time for the treatment group was 2 hours and 2 minutes (2:02, or 122 minutes) with a standard deviation of 22 minutes. rou sse a statistical computing package to conduct the independent-measures t test. The following tables represent your output Group Statistics Std Std Error Group Mean 127 122 Time Control 10 10 29 9.17 6.96

Explanation / Answer

So, here we are required to analyse the given results.

First let us state the thumb rule for determining whether the null hpothesis is to be rejected or not. We do this using p-value of our test. If the p-value is less than prefixed level of significance then we reject the null hypothesis and conclude that there are no significant effect.

Q.1. The null hyptothesis here is the equality of variances. Here we can find the p-value by looking at the "sig" column. Which is .307 and from the question it is clear that the level of significance is .05. Since the p-value is >0.05, we fail to reject the null hypothesis of equal variance. and conclude that the result is not significant meaning that the data do not provide sufficient evidence of variances being different and we can use the "Equal Variance" row. Here the THIRD OPTION is correct.

Q.2. Here the null hypothesis is the equality of means before and after the consumption of the energy bar.

Now, going by the Equal variance row, we find that the p value corresponding to the test is .547. Which is greater than .05. Now, It means that we have to accpet the null hypothesis concluding that there is no evidence that consuming energy bar before 1 hour before the race had an effect on performance t(18)=.647, p=.547

Here also, THIRD OPTION is correct.

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