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 Cognition Researchers from the University of British Columbia conducted a study to investigate the effects of color on cognitive tasks. Words were displayed on a computer screen with background colors of red and blue. Results from scores on a test of word recall are given below. Higher scores correspond to greater word recall.

a. Use a 0.05 significance level to test the claim that the samples are from populations with the same mean.

b. Construct a 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)?

c. Does the background color appear to have an effect on word recall scores? If so, which color appears to be associated with higher word memory recall scores?

Red Background n = 35, x = 15.89, s = 5.90

Blue Background n = 36, x = 12.31, s = 5.48

Short answer

a. There is a significant difference between the means of the two populations at the level of significance $\alpha =0.05$.

b. The 95% confidence interval for this hypothesis test is (0.832,6.33). Thus, the result is similar to part (a).

c. The fact that background color has some effect on word recall scores implies that the samples are from populations with different means. Red is associated with higher word memory recall scores.

Step-by-step solution

Given information

The level of significance and the sample statistics for the test results on samples with red and blue backgrounds are given below.

Level of significance:$\alpha =0.05$

$$\begin{gathered} {\text{1}}^{\text{st}}\text{sample}\text{:Red Background} \\ {n}_1=35, \\ {s}_1=0.590 \\ {\overline{x}}_1=15.89 \end{gathered}$$

$$\begin{gathered} {\text{2}}^{\text{nd}}\text{sample}\text{:Blue Background} \\ {n}_2=36 \\ {s}_2=5.48 \\ {\overline{x}}_2=12.31 \end{gathered}$$

State the hypothesis 

a.

The claim states equality between the means of two independent samples.

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

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

State the test statistic

It is an example of two independent sample t-test concerning means.

The formula for test statistics 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 the critical values

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

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

The critical values are obtained as follows:

$$\begin{gathered} P\left(t>t_{\frac{\alpha }{2}}\right)=\frac{\alpha }{2} \\ P\left(t>t_{\frac{0.05}{2}}\right)=\frac{0.05}{2} \\ P\left(t>t_{0.025}\right)=0.025 \end{gathered}$$

Thus, the critical value obtained from the t-table for 34 degrees of freedom is 2.0322.

Compute the test statistic

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

$$\begin{gathered} t=\frac{\left({\overline{x}}_1-{\overline{x}}_2\right)}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}} \\ =\frac{\left(15.89-12.31\right)}{\sqrt{\frac{{\left(5.90\right)}^2}{35}+\frac{{\left(5.48\right)}^2}{36}}} \\ =2.647 \end{gathered}$$

The test statistic is $t=2.647$.

Decision rule using the critical value

The decision criterion for this problem statement is given below.

If$\left|t_{stat}\right|>t_{crit}$ ; Reject a null hypothesis at $\alpha$level of significance

If $\left|t_{stat}\right|<t_{crit}$ ; Fail to accept the null hypothesis at $\alpha$level of significance

In this case, $\left|t_{stat}=2.647\right|>t_{crit}=2.032$. Thus, the null hypothesis is rejected.

It can be concluded that there is insufficient evidence to support the claim, and hence the two backgrounds have different mean scores.

Step 7:Confidence interval for the difference of means of population

b.

The 0.05 significance level for the two-tailed test implies a confidence level of 95%.

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:

$$\begin{gathered} E=t_{\frac{\alpha }{2}}\times \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}} \\ =t_{\frac{0.05}{2}}\times \sqrt{\frac{{5.90}^2}{35}+\frac{{5.48}^2}{36}} \\ =2.0322\times 1.3523 \\ =2.748 \end{gathered}$$

Substitute the values in the formula to find the confidence interval as follows:

$$\begin{gathered} \text{C.I}=\left({\overline{x}}_1-{\overline{x}}_2\right)-E<\left({\mu }_1-{\mu }_2\right)<\left({\overline{x}}_1-{\overline{x}}_2\right)+E \\ =\left(15.89-12.31\right)-2.748<\left({\mu }_1-{\mu }_2\right)<\left(15.89-12.31\right)+2.748 \\ =0.832<\left({\mu }_1-{\mu }_2\right)<6.33 \end{gathered}$$

Thus, the 95% confidence interval appropriate for the hypothesis test lies between 0.832 and 6.33.

The interval does not include a zero, which implies the null hypothesis is rejected (same as part (a)).

Analyze the results

c.

From the results of the two parts, there is sufficient evidence to warrant the claim that there is a difference in cognitive scores when two different backgrounds are used.

As the interval shows positive values, it is evident that the red background has better scores than the blue background. Thus, the red color is associated with higher word memory recall scores.