In Exercises 5–16, test the given claim.

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.

Red Background:	n = 35, \(\bar x\) = 3.39, s = 0.97
Blue Background:	n = 36, \(\bar x\)= 3.97, s = 0.63

For a sample of 35 subjects who were given a creative task with a red background, the mean score of creativity is equal to 3.39, and the standard deviation is equal to 0.97.For another sample of 36 subjects who were given a creative task with a blue background, the mean score of creativity is equal to 3.97 and the standard deviation is equal to 0.63.

It is claimed that the variation in the creative score withthe blue background is equal to the variation in the creative score withthe red background.

Let\({\sigma _1}\)and\({\sigma _2}\)be the populationstandard deviations of the creative scores corresponding to the red background and the blue background, respectively.

Null Hypothesis:The population standard deviation of the scores with the red background is equal to the population standard deviation of the scores with the blue background.

Symbolically,

\({H_0}:{\sigma _1} = {\sigma _2}\)

Alternative Hypothesis:The population standard deviation of the scores with the red background is not equal to the population standard deviation of the scores with the blue background.

Symbolically,

\({H_1}:{\sigma _1} \ne {\sigma _2}\)

Since two independent samples involve a claim about the population standard deviation, apply an F-test.

Consider the larger sample variance to be\(s_1^2\)and the corresponding sample size to be\({n_1}\).

The following values are obtained:

\({\left( {0.97} \right)^2} = 0.9409\)

\({\left( {0.63} \right)^2} = 0.3969\)

Here,\(s_1^2\)is the sample variance corresponding to the red background and has a value equal to 0.9409.

\(s_2^2\)is the sample variance corresponding to the blue background and has a value equal to 0.3969.

Substitute the respective valuesto calculate the F statistic:

\(\begin{array}{c}F = \frac{{s_1^2}}{{s_2^2}}\\ = \frac{{{{\left( {0.97} \right)}^2}}}{{{{\left( {0.63} \right)}^2}}}\\ = 2.371\end{array}\)

Thus, F is equal to 2.371.

The value of the numerator degrees of freedom is equal to:

\(\begin{array}{c}{n_1} - 1 = 35 - 1\\ = 34\end{array}\)

The value of the denominator degrees of freedom is equal to:

\(\begin{array}{c}{n_2} - 1 = 36 - 1\\ = 35\end{array}\)

For the F test, the critical value corresponding to the right-tail is considered.

The critical value can be obtained using the F-distribution table with numerator degrees of freedom equal to 34 and denominator degrees of freedom equal to 35 for a right-tailed test.

The level of significance is equal to:

\(\begin{array}{c}\frac{\alpha }{2} = \frac{{0.05}}{2}\\ = 0.025\end{array}\)

Thus, the critical value is equal to 1.9678.

The two-tailed p-value for F equal to 2.371 is equal to 0.0129.

Since the test statistic value is greater than the critical value and the p-value is less than 0.05, the null hypothesis is rejected.

Thus, there is enough evidence to rejectthe claimthat creative task scores have the same variation with a red background and a blue background.