In Exercises 5–16, test the given claim.

Color and Recall Researchers from the University of British Columbia conducted trials toinvestigate the effects of color on the accuracy of recall. Subjects were given tasks consistingof words displayed on a computer screen with background colors of red and blue. The subjectsstudied 36 words for 2 minutes, and then they were asked to recall as many of the words as theycould after waiting 20 minutes. Results from scores on the word recall test are given below.

Use a 0.05 significance level to test the claim that the variation of scores is the same with the red background and blue background.

Accuracy Scores

Red Background:	n = 35, \(\bar x\) = 15.89, s = 5.90
Blue Background:	n = 36, \(\bar x\)= 12.31, s = 5.48

For a sample of 35 subjects who were given a taskto recall words with a red background, the mean accuracy score is equal to 15.89, and the standard deviation is equal to 5.90. For another sample of 36 subjects who were given a taskto recall wordswith a blue background, the mean accuracy is equal to 12.31, and the standard deviation is equal to 5.48.

It is claimed that the variation in the accuracy score with the blue background is equal to the variation in the accuracy score with the red background.

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

Null Hypothesis: The populationstandard deviationof the accuracy score with the blue background is equal to the populationstandard deviationof the accuracy score with the red background.

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

Alternative Hypothesis: The populationstandard deviationof the accuracy score with the blue background is not equal to the populationstandard deviationof the accuracy score with the red background.

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

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:

\({\left( {5.90} \right)^2} = 34.81\)

\({\left( {5.48} \right)^2} = 30.0304\)

Here,\(s_1^2\)is the sample variance corresponding to the red background and has a value equal to 34.81.

\(s_2^2\)is the sample variance corresponding to the blue background and has a value equal to 30.0304.

Substitute the respective values to calculate the F statistic:

\(\begin{array}{c}F = \frac{{s_1^2}}{{s_2^2}}\\ = \frac{{{{\left( {5.90} \right)}^2}}}{{{{\left( {5.48} \right)}^2}}}\\ = 1.159\end{array}\)

Thus, F is equal to 1.159.

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 1.159 is equal to 0.6659.

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

Thus, there is not enough evidence to rejectthe claimthat accuracy scores have the same variation with a red background and a blue background.