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Test the given claim. 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. $$\begin{array}{l|l} \hline \text { Red Background: } & n=35, \bar{x}=3.39, s=0.97 \\ \hline \text { Blue Background: } & n=36, \bar{x}=3.97, s=0.63 \\ \hline \end{array}$$

Short Answer

Expert verified
Reject the null hypothesis. There is significant evidence to suggest differences in creativity score variances between red and blue backgrounds.

Step by step solution

01

State the Hypotheses

Formulate the null hypothesis \(H_0\) and the alternative hypothesis \(H_a\). For this exercise, the null hypothesis is \(H_0: \sigma_{Red}^2 = \sigma_{Blue}^2\) which states that the variances are equal. The alternative hypothesis is \(H_a: \sigma_{Red}^2 eq \sigma_{Blue}^2\), indicating the variances are not equal.
02

Identify the Test Statistic

Use the F-test for comparing two variances. The test statistic is \[ F = \frac{s_{Red}^2}{s_{Blue}^2} \] where \( s_{Red}^2 \) and \( s_{Blue}^2 \) are the sample variances for the red and blue backgrounds respectively.
03

Compute the Test Statistic

Calculate the variances for both groups from the given standard deviations: \[ s_{Red}^2 = (0.97)^2 = 0.9409 \] \[ s_{Blue}^2 = (0.63)^2 = 0.3969 \] Now, compute the F-test statistic: \[ F = \frac{0.9409}{0.3969} \] \[ F \approx 2.37 \]
04

Determine the Critical Value

The degrees of freedom for the numerator (Red background) are \( df_1 = n_{Red} - 1 = 34 \), and for the denominator (Blue background) \( df_2 = n_{Blue} - 1 = 35 \). Using an F-table or F-distribution calculator at \( \alpha = 0.05 \), find the critical value for \( df_1 = 34 \) and \( df_2 = 35 \). The critical values are approximately \ F_{lower} = 0.48 \ and \ F_{upper} = 2.20 \.
05

Compare the Test Statistic to the Critical Value

Compare the calculated F-test statistic (2.37) to the critical values (0.48 and 2.20). Since \ F \approx 2.37 \ is greater than \ F_{upper} \ (2.20), we reject \( H_0 \).
06

State the Conclusion

Based on the comparison, we reject the null hypothesis. This means there is significant evidence to suggest that the variances in creativity task scores differ between the red and blue backgrounds.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

F-test
The F-test is a statistical method used to compare the variances of two different populations or groups. When we talk about variances, we are discussing how spread out or how varied the data points are in each group. The F-test essentially tells us if the difference in variability between two groups is significant or not.

In our example with the red and blue backgrounds, we want to see if the variability in creativity scores is different between the two groups. To do this, we compute the test statistic using the formula: \[ F = \frac{s_{Red}^2}{s_{Blue}^2} \] Here, \( s_{Red}^2\) and \( s_{Blue}^2\) are the variances for the red and blue groups, respectively.

The resulting F-value can then be compared to a critical value from the F-distribution table to determine if the variances are significantly different.
Variance Comparison
Comparing variances helps us understand the consistency of data within different groups. High variance means the data points are spread out, while low variance means they are closer to the mean.

In our creativity study, we found: \[ s_{Red}^2 = 0.9409 \] \[ s_{Blue}^2 = 0.3969 \]
These values represent how varied the scores are for each background color.

By comparing these variances using the F-test, we learn if the difference in spread is statistically significant. Such comparisons are essential in experiments because they help us determine if observed differences are due to the experimental conditions or just random variations.
Null Hypothesis
The null hypothesis (\( H_0 \) ) is a statement that assumes no effect or no difference. It is the default position that we test against.

In the given exercise, our null hypothesis is that the variances are equal: \[ H_0: \sigma_{Red}^2 = \sigma_{Blue}^2 \]
This means we initially assume that the background color does not affect the variation in creativity scores. The alternative hypothesis (\( H_a \)), on the other hand, suggests that the variances are not equal: \[ H_a: \sigma_{Red}^2 \e \sigma_{Blue}^2 \]
If our test provides enough evidence to reject \( H_0 \), we conclude that the red and blue backgrounds result in different levels of variability in creativity scores.
Significance Level
The significance level (\( \alpha \) ) helps us decide how strong the evidence must be to reject the null hypothesis. It represents the probability of rejecting \( H_0 \) when it is actually true.

A common choice is 0.05 (or 5%), meaning we are willing to accept a 5% risk of wrongly rejecting \( H_0 \).

In our study, we use \( \alpha = 0.05 \) when testing the variances of creativity scores. After calculating the test statistic (F-value), we compare it against critical values from the F-distribution table specific to this significance level.

If our F-value falls outside the acceptance range (bounded by \( F_{lower}\) and \( F_{upper}\) critical values), we reject \( H_0 \), indicating that the variances differ significantly.

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Most popular questions from this chapter

Test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value or critical value(s), then state the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim. The Chapter Problem involved passenger cars in Connecticut and passenger cars in New York, but here we consider passenger cars and commercial trucks. Among 2049 Connecticut passenger cars, 239 had only rear license plates. Among 334 Connecticut trucks, 45 had only rear license plates (based on samples collected by the author). A reasonable hypothesis is that passenger car owners violate license plate laws at a higher rate than owners of commercial trucks. Use a 0.05 significance level to test that hypothesis. a. Test the claim using a hypothesis test. b. Test the claim by constructing an appropriate confidence interval.

Test the given claim. Blanking Out on Tests Many students have had the unpleasant experience of panicking on a test because the first question was exceptionally difficult. The arrangement of test items was studied for its effect on anxiety. The following scores are measures of "debilitating test anxiety" which most of us call panic or blanking out (based on data from "Item Arrangement, Cognitive Entry Characteristics, Sex and Test Anxiety as Predictors of Achievement in Examination Performance,"by Klimko, Joumal of Experimental Education, Vol. 52, No. 4.) Using a 0.05 significance level, test the claim that the two populations of scores have different amounts of variation. $$\begin{array}{|c|c|c|c|c|} \hline \multicolumn{5}{|c|} {\text { Questions Arranged from Easy to Difficult }} \\ \hline 24.64 & 39.29 & 16.32 & 32.83 & 28.02 \\ \hline 33.31 & 20.60 & 21.13 & 26.69 & 28.90 \\ \hline 26.43 & 24.23 & 7.10 & 32.86 & 21.06 \\ \hline 28.89 & 28.71 & 31.73 & 30.02 & 21.96 \\ \hline 25.49 & 38.81 & 27.85 & 30.29 & 30.72 \\ \hline \end{array}$$ $$\begin{array}{|c|c|c|c|} \hline \multicolumn{3}{|c|} {\text { Questions Arranged from Difficult to Easy }} \\ \hline 33.62 & 34.02 & 26.63 & 30.26 \\ \hline 35.91 & 26.68 & 29.49 & 35.32 \\ \hline 27.24 & 32.34 & 29.34 & 33.53 \\ \hline 27.62 & 42.91 & 30.20 & 32.54 \\ \hline \end{array}$$

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 \(\boldsymbol{n}_{I}-\boldsymbol{I}\) and \(\boldsymbol{n}_{2}-\boldsymbol{I} .\) ) Blanking Out on Tests Many students have had the unpleasant experience of panicking on a test because the first question was exceptionally difficult. The arrangement of test items was studied for its effect on anxiety. The following scores are measures of "debilitating test anxiety." which most of us call panic or blanking out (based on data from "Item Arrangement, Cognitive Entry Characteristics, Sex and Test Anxiety as Predictors of Achievement in Examination Performance," by Klimko, Journal of Experimental Education, Vol. 52, No. 4.) Is there sufficient evidence to support the claim that the two populations of scores have different means? Is there sufficient evidence to support the claim that the arrangement of the test items has an effect on the score? Is the conclusion affected by whether the significance level is 0.05 or 0.01? $$\begin{array}{|c|c|c|c|c|} \hline {}{} {\text { Questions Arranged from Easy to Difficult }} \\ \hline 24.64 & 39.29 & 16.32 & 32.83 & 28.02 \\ \hline 33.31 & 20.60 & 21.13 & 26.69 & 28.90 \\ \hline 26.43 & 24.23 & 7.10 & 32.86 & 21.06 \\ \hline 28.89 & 28.71 & 31.73 & 30.02 & 21.96 \\ \hline 25.49 & 38.81 & 27.85 & 30.29 & 30.72 \\ \hline\end{array}$$ $$\begin{array}{|c|c|c|c|} \hline {}{} {\text { Questions Arranged from Difficult to Easy }} \\ \hline 33.62 & 34.02 & 26.63 & 30.26 \\ \hline 35.91 & 26.68 & 29.49 & 35.32 \\ \hline 27.24 & 32.34 & 29.34 & 33.53 \\ \hline 27.62 & 42.91 & 30.20 & 32.54 \\ \hline\end{array}$$

Test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value or critical value(s), then state the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim. A trial was conducted with 75 women in China given a 100-yuan bill, while another 75 women in China were given 100 yuan in the form of smaller bills (a 50-yuan bill plus two 20-yuan bills plus two 5-yuan bills). Among those given the single bill, 60 spent some or all of the money. Among those given the smaller bills, 68 spent some or all of the money (based on data from "The Denomination Effect," by Raghubir and Srivastava, Journal of Consumer Research, Vol. 36). We want to use a 0.05 significance level to test the claim that when given a single large bill, a smaller proportion of women in China spend some or all of the money when compared to the proportion of women in China given the same amount in smaller bills. a. Test the claim using a hypothesis test. b. Test the claim by constructing an appropriate confidence interval. c. If the significance level is changed to 0.01, does the conclusion change?

Test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value or critical value(s), then state the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim. In a study of Burger King drive-through orders, it was found that 264 orders were accurate and 54 were not accurate. For McDonald's, 329 orders were found to be accurate while 33 orders were not accurate (based on data from \(Q S R\) magazine). Use a 0.05 significance level to test the claim that Burger King and McDonald's have the same accuracy rates. a. Test the claim using a hypothesis test. b. Test the claim by constructing an appropriate confidence interval. c. Relative to accuracy of orders, does either restaurant chain appear to be better?

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