In Exercises 5–20, assume that the two samples are independent...

Question

In Exercises 5–20, assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. (Note: Answers in Appendix D include technology answers based on Formula 9-1 along with “Table” answers based on Table A-3 with df equal to the smaller of n1-1 and n2-1)

Color and Cognition Researchers from the University of British Columbia conducted a study to investigate the effects of color on cognitive tasks. Words were displayed on a computer screen with background colors of red and blue. Results from scores on a test of word recall are given below. Higher scores correspond to greater word recall.

a. Use a 0.05 significance level to test the claim that the samples are from populations with the same mean.

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Accepted Answer

Welcome back, everyone. In this problem, a nutritionist wants to compare the effects of two different diets and weight loss. After 8 weeks, the following statistics are recorded for the sample size, mean weight loss, and the standard deviation of diet X and the diet Y. Assume that the two samples are independent simple random samples selected from normally distributed populations and do not assume that the population standard deviations are equal. Use degrees of freedom equal to the smaller of N1 minus 1 and N2 m1. At the 0.05 significance level, test the claim that there is no significant difference in mean weight loss between the two diets. Now, first, we need to state our hypotheses to carry out the test. Like our problem statement tells us, we're testing the claim that there is no significant difference in mean weight loss between the two diets, OK? So our another hypothesis is that there is no difference, that the first mean equals the second mean. And our alternative hypothesis is that there is a difference. In other words, the first mean is not equal to the second mean. Now, if we think about the information that we have here, OK, then so far we know, let me just separate these. We know that for diet X. X bar 1, OK, if we consider it the first group, X bar 1 is equal to 8.2. OK. Uh, S1 is equal to 2.4, and N1, OK, is equal to 28. On the other hand, for diet why. We know that X2, considering it to be the second group, is going to be 6.5, S2 is 2.9, and N2 is 30. Now we can test or claim by using the two sample T test for unequal variances. So what are we really trying to find? Well, like our problems says that we are assuming that the two samples are selected from normally distributed populations. So if we draw a normal distribution, OK, and we have our critical values. We know that this is a T test since we're testing if one value is equal to the other. So if we find our critical values here, or criticalt values, if our test statistic falls outside of the critical values, OK, then we can reject the null hypothesis. But if it falls within our critical values, then we cannot reject. We fail to reject, uh, the null hypothesis. In other words, there's not enough evidence that would be there to support the claim. So there are two things we need. First, we need our critical values, and then we need to calculate the test statistic. OK, let's call the STC and then figure out where the test statistic lies. Now let's start by calculating the test statistic, OK? And let me put that in red. Calculating. Calculating the test statistic. Now to do that, then we're gonna solve for T and recall that it equals the difference between the sample means X 1 minus X2 minus mu1 minus m2, which is often assumed to be zero, OK, all divided by the square root, OK, of S1 2 divided by N1 plus S2 2 divided by N2. Now in this case, if we substitute our known values, this is gonna be 8.2 minus 6.5 minus 0. All divided by the square root, OK, of 2.4 squared divided by 28 plus 2.9 squared divided by 30, and in this case we should get our test statistic then to be equal to 2.438. Know that we have the test test test statistic, then we need to find the critical value to determine where it lies. Now, if we think about it here, like we were told in our problem statement, we're gonna use degrees of freedom equal to the smaller of N1 minus 1 and N2 m1 and we're at the 0.05 um significance level, OK? So for a two-tailed test. Which is what we're testing for, so for a two-tailed test. At alpha equals 0.05 and or degrees of freedom to be equal to the minimum value of 28 minus 1 and 30 minus 1, which would be 27 and 29, so the minimum is 27, that's the smaller of both those values. Then are the criticality value, OK. The critical T value, which is going to be a half of alpha, which is 0.025 and our degrees of freedom 27. By using our table of T values is going to be approximately equal to 2.052. So now, for that criticality value, we can make our decision. If we go back to our diagram here, then we can replace our critical values on this diagram with negative 2.052 on the left side and 2.052 on the right side. And no, the question is, where does our test statistic fall? Well, 2.438 is greater than 2.052, so it would fall within the rejection region, OK? In other words, here, if I just make some space, we know that it would probably be somewhere here. OK? And know since it falls in the rejection region, therefore we reject the no hypothesis. So, to in our concluding statement, that means that at the 0.05 significance level, OK, let's write that out. There is sufficient evidence. There is sufficient evidence. To conclude That there is. A significant difference. OK. And mean weight loss. OK. Let's just make our statement here and mean weight loss. Between the two diets. OK? We can make that claim. Thanks a lot for watching everyone. I hope this video helped.

Key Concepts

Independent Samples

Hypothesis Testing

t-Test for Independent Samples

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