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 Creativity Researchers from the University of British Columbia conducted trials to investigate the effects of color on creativity. Subjects with a red background were asked to think of creative uses for a brick; other subjects with a blue background were given the same task. Responses were scored by a panel of judges and results from scores of creativity are given below. Higher scores correspond to more creativity. The researchers make the claim that “blue enhances performance on a creative task.”

a. Use a 0.01 significance level to test the claim that blue enhances performance on a creative task.

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Accepted Answer

Welcome back, everyone. In this problem, researchers are studying the effect of background music on concentration. One group of students studied with classical music while another group studied in silence. Their concentration scores were recorded and summarized as follows. For the classical music group, the sample size was 35, the mean was 7.2, and the standard deviation was 1.1, while the group in silence had a sample size of 38, a mean of 65, and a standard deviation of 1.4. 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 minus 1. At the 0.01 significance level, test the claim that classical music leads to higher concentration scores than silence. Now, in this case, we are testing the claim, OK, like we said, that classical music leads to higher concentration scores than silence, OK? So that means then. If we let m one be the mean for uh the classical music, while mu 2 represent the mean for silence, then our another hypothesis is that mu 1 equals mu 2, there is no difference, which makes our alternative hypothesis be that mu1 is greater than mu 2, that classical music leads to higher concentration scores than silence, OK? So, no, if we identify the information that we have so far, if we say that classical music is group one, then for classical music, X bar 1 is 7.2, S1 is 1.1, and N1 is 35. On the other hand, if we make silence a second group, X2 equals 6.5. S2 equals 1.4, OK, and N2 equals 38. So, no, in that case, that means we'll need to solve. How can we do that? Well, in this case, because it is a one-tailed test and we're using a two samplete test, OK? Then we can set up our normal distribution here, assuming, like we said, it's selected from normally distributed populations. And what we're really trying to find is that we want to know if For our criticalt value, we want to find a criticalt value and figure out if it's greater than or the test statistic is greater than that criticalt value. If it's greater, then we reject based on the one tail test, and if it's not, then there's not enough evidence for us to reject the null hypothesis. So we'll need a criticalt value and the test statistic. How can we get both of those? Well, let's start by finding the criticality value, OK? Now for a one tailed test, OK, like we said earlier, that's what we're using. At a significant level of 0.01 and. The degrees of freedom being the smaller of the two samples minus 1. So in this case it's the smaller of 35 minus 1, which is 34. And 38 minus 1, which is 37, so the smaller value is 34. So we're gonna use 34 degrees of freedom. Then from our tea tables, the critical T value is going to be approximately equal. The 2.441. So now we need to test if or figure out if the test statistic is greater than equal to or less than that value. How can we calculate the test statistic? We recall that for the two sample T test with unequal variances, the test statistic will be equal to the difference between the sample means. Minus 1 minus 2, OK, where mu 1 minus m2 is often assumed to be zero, divided by the square root. Of S1 square divided by N1. Plus S2 squad divided by N2. If we substitute our known values, then that's going to be 7.2 minus 6.5 minus 0, all divided by the square root, OK, of 1.1 square divided by 35 plus 1.4 square divided by 38. And if we go ahead and solve here, then we should get the test statistic to be approximately 2.385. Now how does this test statistic compare with our critical value? Well, if we think about it on our distribution, that means it will probably fall somewhere that's a little bit before our critical value. OK. So we could say that it probably falls somewhere right here, OK? It's 2.385. And since it does not fall after it, that means we cannot reject the null hypothesis. There's not enough evidence for us to do that. So since 2.385 is less than our critical value of 2.441. We fail to reject the null hypothesis, so what can we conclude? Well, in that case, at the 0.01 significance level. Yeah? This means that there is not enough evidence. Now we're we're putting the statistics giving meaning based on our scenario to support the claim. OK, so we're looking at our context. That classical music. A music leads to. Higher concentration squares. Higher concentration scores. Than silence. OK. So this could be the conclusion we could make based on the information we found. Thanks a lot for watching everyone. I hope this video helped.

Key Concepts

Hypothesis Testing

Significance Level

Independent Samples t-Test

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