91影视

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:$伪=0.05$

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

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

a.

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

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

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

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

The formula for test statistics 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 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:

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

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

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

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

The test statistic is $t=2.647$.

The decision criterion for this problem statement is given below.

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

If $\left|t_{stat}\right|<t_{crit}$ ; Fail to accept the null hypothesis at $伪$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.

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

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

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

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

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

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

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)).

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.