The claim attempts to test if a blue background enhances creativity scores at a 0.01 level of significance.

\(\begin{array}{l}{{\rm{1}}^{{\rm{st}}}}\;{\rm{sample}}\;{\rm{:Red Background}}\\{n_1}\; = 35,\\{s_1} = 0.97\\{{\bar x}_1} = 3.39\end{array}\)

\(\begin{array}{l}{{\rm{2}}^{{\rm{nd}}}}\;{\rm{sample}}\;{\rm{:Blue Background}}\\{n_2} = 36\\{s_2} = 0.63\;\\{{\bar x}_2} = 3.97\end{array}\)

a.

As per the claim, the hypotheses are formulated as follows:

\(\begin{array}{l}{H_{0\;}}:{\mu _1} = {\mu _2}\\{H_1}\;:{\mu _1} < \;{\mu _2}\end{array}\)

Here,\({\mu _1},{\mu _2}\)are the population means of creativity scores for red background and blue background, respectively.

The test is left tailed.

This is an example of two independent samples t-test about means.

The formula for test-statistic is given below.

\(t = \frac{{\left( {{{\bar x}_1} - {{\bar x}_2}} \right) - \left( {{\mu _1} - {\mu _2}} \right)}}{{\sqrt {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} }}\)

For t-distribution, find degrees of freedom as follows:

\(\begin{array}{c}df = \min \left( {\left( {{n_1} - 1} \right),\left( {{n_2} - 1} \right)} \right)\\ = \min \left( {\left( {35 - 1} \right),\left( {36 - 1} \right)} \right)\\ = 34\end{array}\)

For a left tailed test, the critical values are obtained as follows:

\(\begin{array}{c}P\left( {t < {t_\alpha }} \right) = \alpha \\P\left( {t < {t_{0.01}}} \right) = 0.01\end{array}\)

Thus, the critical value obtained from the t-table for 34 degrees of freedom is -2.441.

The test statistic of the means of populations is as follows:

\(\begin{array}{c}{t_{stat}} = \frac{{\left( {{{\bar x}_1} - {{\bar x}_2}} \right)}}{{\sqrt {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} }}\\ = \frac{{\left( {3.39 - 3.97} \right)}}{{\sqrt {\frac{{{{\left( {0.97} \right)}^2}}}{{35}} + \frac{{{{\left( {0.63} \right)}^2}}}{{36}}} }}\\ = - 2.979\end{array}\) ;

The test statistic is \(t = - 2.987\).

The decision criterion for this problem statement is given below.

If thetest statistic is lesser than the critical value, reject the null hypothesis at\({\bf{\alpha }}\)level of significance.

If the test statistic is greater than the critical value, fail to accept the null hypothesis at\({\bf{\alpha }}\)level of significance.

In this case, \( - 2.979 < - 2.441\). Thus, the null hypothesis is rejected.

It shows that there is enough evidence to support the claim that students with a red background were less creative than students with a blue background.

b.

The confidence level corresponding to the 0.01 level of significance for a one-tailed test is 98%.

The formula for the confidence interval of the means of population is given by

\(\left( {{{\bar x}_1} - {{\bar x}_2}} \right) - E < \left( {{\mu _1} - {\mu _2}} \right) < \left( {{{\bar x}_1} - {{\bar x}_2}} \right) + E\)

E is the margin of error and the formula for the margin of error is as follows:

\(\begin{array}{c}E = {t_{\frac{\alpha }{2}}} \times \sqrt {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} \\ = {t_{\frac{{0.02}}{2}}} \times \sqrt {\frac{{{{0.97}^2}}}{{35}} + \frac{{{{0.63}^2}}}{{36}}} \\ = 2.441 \times 0.1947\\ = 0.475\end{array}\)

Substitute all derived values in the formula and find the confidence interval.

\(\begin{array}{c}{\rm{C}}{\rm{.I}} = \left( {{{\bar x}_1} - {{\bar x}_2}} \right) - E < \left( {{\mu _1} - {\mu _2}} \right) < \left( {{{\bar x}_1} - {{\bar x}_2}} \right) + E\\ = \left( {3.39 - 3.97} \right) - 0.475 < \left( {{\mu _1} - {\mu _2}} \right) < \left( {3.39 - 3.97} \right) + 0.475\\ = - 1.06 < \left( {{\mu _1} - {\mu _2}} \right) < - 0.10\end{array}\)

The confidence interval of 98% lies between -1.06 and -0.10.

The interval does not include 0; so there is enough evidence to support the claim that the mean creativity score with blue background is greater than red background.

Thus, it implies that blue enhances creativity score.