Color and Creativity Researchers from the University of Britis...

Question

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. Use a 0.05 significance level to test the claim that creative task scores have the same variation with a red background and a blue background.
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Accepted Answer

Hello everyone. Let's take a look at this question together. A psychologist is studying the effect of background music on concentration levels. One group of students took a test with classical music playing in the background, while another group took the same test with no music. The scores were analyzed for variability. The results are as follows. Classical music has an end value of 28, an X bar of 75.2, and an S of 8.1. And no music has an N value of 30, an X bar of 78.5, and an S of 5.6. At the 0.05 significance level, test the claim that the variances of test scores are the same for both groups. Assume that the samples are from normally distributed populations. So the first step in solving this problem is to do our requirement check, starting with the first requirement, which is the two populations are independent. As the data collected from one group does not influence or relate to the data from the other group. And the second requirement point is assume that the two samples are simple random samples, and that the psychologist carefully planned the design for the study so that the selection process follows a random procedure. And the last point for the requirement check is that each of the two populations is assumed to be normally distributed. So the next step in solving this problem is to define the hypotheses, starting with the null hypothesis. Which states that population variance 1 is equal to population variance 2, and then we have our alternative hypothesis, which states that population variance 1 does not equal population variance 2. And then our next step is to calculate the test statistic F, which can be calculated by the formula F is equal to the sample variance 1 divided by the sample variance 2. Which determining whether the classical music or no music is sample variance 1 or 2, we look at the values as the sample variance 1 is the larger variance. And from the data in the question, we see classical music as a sample variance of 8.1, whereas no music has a sample variance of 5.6. So classical music with the larger sample variants will be our first group. And from the data in the problem, we note that group one, which is the classical music, has the sample variance of 8.1 and an end value of 28, and group two, which is no music, has the sample variance of 5.6, and an end value of 30. So plugging in the values for sample variants 1 and 2, our formula for F is equal to 8.1 squared divided by 5. 5.6 squared, resulting in 2.092 as the value for F. And then our next step is to calculate the degrees of freedom for both groups, which is calculated by N minus 1. So for the first group, we know that our N value is 28, so 28 subtracted by 1 is equal to 27, so 27 is our first degree of freedom number. And then we calculate the same for our second degree of freedom number using the N2 value subtracted by 1, which is 30 minus 1, giving us 29. So 29 is our second degree of freedom number. And using the degrees of freedom, we can find the critical value where we look up the upper critical value for the calculated degrees of freedom, and for alpha divided by 2 is equal to 0.025. So we use the table for F distribution, and since the first degree of freedom, which is equal to 27, is between 24 and 30, we perform linear interpolation, and the upper critical value will be represented as F sub upper. So by linear interpolation, we have F upper minus. 2.1540 divided by 27 minus 24 is equal to 2.0923 minus 2.1540 divided by 30 minus 24, which we are trying to find the value for F upper, so we rearrange the formula to give us F upper is in parentheses 2.0923 minus 2.1540 divided by 30 minus 24 multiplied by in parentheses 27 minus 24 plus 2.1540, which results in 2.123 as the value for F stub upper. Therefore, the upper critical value, which is F sub upper, is 2.213. Therefore, if F is greater than 2.123, we reject the null hypothesis, and we know that 2.092 is less than 2.1. 3, we fail to reject the null hypothesis. So in conclusion, at the 0.05 significance level, there is insufficient evidence to warrant rejection of the claim that the variances of test scores are the same for both groups. I hope you found this video to be helpful. Thank you and goodbye.

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Key Concepts

Hypothesis Testing

Significance Level

Variance and Comparison of Variances

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