91影视

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

$$\begin{gathered} {\text{1}}^{\text{st}}\text{sample}\text{:Red Background} \\ {n}_1=35, \\ {s}_1=0.97 \\ {\stackrel{炉}{x}}_1=3.39 \end{gathered}$$

$$\begin{gathered} {\text{2}}^{\text{nd}}\text{sample}\text{:Blue Background} \\ {n}_2=36 \\ {s}_2=0.63 \\ {\stackrel{炉}{x}}_2=3.97 \end{gathered}$$

a.

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

$$\begin{gathered} H_0:{渭}_1={渭}_2 \\ {H}_1:{渭}_1<{渭}_2 \end{gathered}$$

Here, ${渭}_1,{渭}_2$are the population means of creativity scores for red background and blue background, respectively.

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

The formula for test-statistic is given below.

$t=\frac{\left({\stackrel{炉}{x}}_1-{\stackrel{炉}{x}}_2\right)-\left({渭}_1-{渭}_2\right)}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}$

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

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

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

$$\begin{gathered} P\left(t<t_{伪}\right)=伪 \\ P\left(t<t_{0.01}\right)=0.01 \end{gathered}$$

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:

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

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 level of significance.

If the test statistic is greater than the critical value, fail to accept the null hypothesis at 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

$\left({\stackrel{炉}{x}}_1-{\stackrel{炉}{x}}_2\right)-E<\left({渭}_1-{渭}_2\right)<\left({\stackrel{炉}{x}}_1-{\stackrel{炉}{x}}_2\right)+E$

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

$$\begin{gathered} E=t_{\frac{伪}{2}}脳\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}} \\ =t_{\frac{0.02}{2}}脳\sqrt{\frac{{0.97}^2}{35}+\frac{{0.63}^2}{36}} \\ =2.441脳0.1947 \\ =0.475 \end{gathered}$$

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

$$\begin{gathered} \text{C.I}=\left({\stackrel{炉}{x}}_1-{\stackrel{炉}{x}}_2\right)-E<\left({渭}_1-{渭}_2\right)<\left({\stackrel{炉}{x}}_1-{\stackrel{炉}{x}}_2\right)+E \\ =\left(3.39-3.97\right)-0.475<\left({渭}_1-{渭}_2\right)<\left(3.39-3.97\right)+0.475 \\ =-1.06<\left({渭}_1-{渭}_2\right)<-0.10 \end{gathered}$$

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.