Question

Independent random samples from normal populations produced the results shown in the next table.

Sample 1	Sample 2
$$\begin{gathered} \mathbf{1}\mathbf{.}\mathbf{2} \\ \mathbf{3}\mathbf{.}\mathbf{1} \\ \mathbf{1}\mathbf{.}\mathbf{7} \\ \mathbf{2}\mathbf{.}\mathbf{8} \\ \mathbf{3}\mathbf{.}\mathbf{0} \end{gathered}$$	$$\begin{gathered} \mathbf{4}\mathbf{.}\mathbf{2} \\ \mathbf{2}\mathbf{.}\mathbf{7} \\ \mathbf{3}\mathbf{.}\mathbf{6} \\ \mathbf{3}\mathbf{.}\mathbf{9} \end{gathered}$$

a. Calculate the pooled estimate of ${\text{σ}}^{\text{2}}$.

b. Do the data provide sufficient evidence to indicate that ${\text{μ}}_{\text{2}}{\text{>μ}}_{\text{1}}$? Test using $\text{α=.10}$.

c. Find a $90\%$ confidence interval for $({\text{μ}}_{\text{1}}-{\text{μ}}_{\text{2}})$.

d. Which of the two inferential procedures, the test of hypothesis in part b or the confidence interval in part c, provides more information about $({\text{μ}}_{\text{1}}-{\text{μ}}_{\text{2}})$?

Short answer

The pooled variance is a rough approximation of the shared variance.

Step-by-step solution

Step-by-Step Solution Step 1: Definition of the pooled estimator.

The pooled estimatoris one that is derived by merging data from two or more separate samples from groups that are thought to have a similar mean. The pooled variance is a technique for estimating common variance.

The formula to find pooled estimator of the variance of two samples is:

${{\mathbf{\text{s}}}_{\mathbf{\text{p}}}}^{\mathbf{\text{2}}}\mathbf{\text{=}}\frac{\mathbf{(}{\mathbf{\text{n}}}_{\mathbf{\text{1}}}\mathbf{-}\mathbf{\text{1}}\mathbf{)}{{\mathbf{\text{s}}}_{\mathbf{\text{1}}}}^{\mathbf{\text{2}}}\mathbf{\text{+}}\mathbf{(}{\mathbf{\text{n}}}_{\mathbf{\text{2}}}\mathbf{-}\mathbf{\text{1}}\mathbf{)}{{\mathbf{\text{s}}}_{\mathbf{\text{2}}}}^{\mathbf{\text{2}}}}{{\mathbf{\text{n}}}_{\mathbf{\text{1}}}{\mathbf{\text{+n}}}_{\mathbf{\text{2}}}\mathbf{-}\mathbf{\text{2}}}$

(a) Calculate a pooled estimate of variance.

Mean of sample 1 = ${\overline{x}}_1=\frac{1.2+3.1+1.7+2.8+3.0}{5}=2.36$

role="math" localid="1652802728137" $\begin{array}{l}{s_1}^2=\frac{1}{n_1-1}\sum _{i=1}^n{[x_i-\overline{x}]}^2\\ =\frac{1}{5-1}[{(1.2-2.36)}^2+{(3.1-2.36)}^2+{(1.7-2.36)}^2+{(2.8-2.36)}^2+{(3.0-2.36)}^2]\\ =\frac{1}{4}[1.3456+0.5476+0.4356+0.1936+0.4096]\\ =\frac{1}{4}\left(2.932\right)\\ =0.733\end{array}$

Mean of sample 2 = ${\overline{x}}_2=\frac{4.2+2.7+3.6+3.9}{4}=3.6$

$\begin{array}{l}{s_2}^2=\frac{1}{n_2-1}\sum _{i=1}^n{[x_i-\overline{x}]}^2\\ =\frac{1}{4-1}[{(4.2-3.6)}^2+{(2.7-3.6)}^2+{(3.6-3.6)}^2+{(3.9-3.6)}^2]\\ =\frac{1}{3}[0.36+0.81+0+0.09]\\ =\frac{1}{3}\left(1.26\right)\\ =0.42\end{array}$

$\begin{array}{l}{s_p}^2=\frac{(5-1)0.733+(4-1)0.42}{5+4-2}\\ =\frac{2.932+1.26}{7}\\ =0.6\end{array}$

Therefore, the pooled estimate of variance is 0.6

(b) Conduct a t-test.

Null Hypothesis,${\text{H}}_{\text{0}}{\text{:μ}}_{\text{1}}\ge {\text{μ}}_{\text{2}}$and

Alternate Hypothesis,${\text{H}}_{\text{a}}{\text{:μ}}_{\text{1}}{\text{<μ}}_{\text{2}}$

The level of significance is $0.10$.

$\begin{array}{c}\mathbf{Degree}\mathbf{}\mathbf{of}\mathbf{}\mathbf{freedom}=n_1+n_2-2\\ =5+4-2\\ =7\end{array}$

From the t-distribution table, the critical value at $0.10$the level of the significance for degrees of freedom about the right-tailed test is $-1.415$.

$\begin{array}{l}t=\frac{\overline{x_1}-{\overline{x}}_2}{\sqrt{{s_p}^2\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}}\\ =\frac{2.36-3.6}{\sqrt{0.6\left(\frac{1}{5}+\frac{1}{4}\right)}}\\ =\frac{-1.24}{\sqrt{0.6(0.2+0.25)}}\\ =\frac{-1.24}{0.52}\\ =-2.38\end{array}$

As, the value of $t<-1.415$, the null hypothesis should be rejected.

Therefore, the data provide sufficient evidence to indicate that ${\mu }_2>{\mu }_1$.

(c) Find confidence interval.

The 90% confidence interval for the difference in means

$\begin{array}{l}=({\overline{x}}_1-{\overline{x}}_2)\pm t_{\alpha /2}\times \sqrt{\frac{{s_1}^2}{n_1}+\frac{{s_2}^2}{n_2}}\\ =(2.36-3.6)\pm 1.895\times \sqrt{\frac{0.733}{5}+\frac{0.42}{4}}\\ =(-1.24)\pm (1.895\times 0.502)\\ =-1.24\pm 0.95\end{array}$

Therefore, the confidence interval for the difference of means is$-2.19\text{to}-0.29$

(d) State the conclusion.

The confidence interval tells us the specific limit within which the difference between the population means is expected to lie with 90% confidence, whereas the hypothesis testing presents the situation where we can tell that ${\mu }_2>{\mu }_1$without specifying any value of the difference between the population means.

Therefore, the confidence interval provides more information ${\mu }_1-{\mu }_2$.