Question

In Exercises 5–20, 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 n1−1 and n2−1.)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. Higher scores correspond to more creativity. The researchers make the claim that “blue enhances performance on a creative task.”

a. Use a 0.01 significance level to test the claim that blue enhances performance on a creative task. b. Construct the confidence interval appropriate for the hypothesis test in part (a). What is it about the confidence interval that causes us to reach the same conclusion from part (a)?Red Background: n = 35, x = 3.39, s = 0.97Blue Background: n = 36, x = 3.97, s = 0.63

Short answer

a. There is sufficient evidence that students with a blue background have performed better as compared to students with a red background.

b. The confidence interval for the difference between the means of the sample is (-1.11,-0.049). As 0 is not included in the interval, the result is supportive of the claim of this test. Therefore, the confidence interval also gives the same conclusion as part (a).

Step-by-step solution

Given information

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

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

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

State the hypothesis 

a.

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

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

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

Compute the test statistic

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

The formula for test-statistic is given below.

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

Find degrees of freedom and critical value

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:

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

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

Compute the test statistic

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

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

The test statistic is $t=-2.987$.

State the decision rule using the critical value

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.

Confidence interval for the difference of means of population

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

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

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

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

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

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

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

Conclude the results from confidence interval

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.