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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:

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

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

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

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

$\begin{array}{l}{s_2}^2=\frac{1}{n_2−1}\underset{i=1}{\overset{n}{∑}}{[x_i−\stackrel{¯}{x}]}^2\\ =\frac{1}{4−1}[{(4.2−3.6)}^2+{(2.7−3.6)}^2+{(3.6−3.6)}^2+{(3.9−3.6)}^2]\\ =\frac{1}{3}[0.36+0.81+0+0.09]\\ =\frac{1}{3}\left(1.26\right)\\ =0.42\end{array}$

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

Therefore, the pooled estimate of variance is 0.6

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

Alternate Hypothesis,${\text{H}}_{\text{a}}{\text{:μ}}_{\text{1}}{\text{\&±ô³Ù;μ}}_{\text{2}}$

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 $-1.415$.

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

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

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

The 90% confidence interval for the difference in means

$\begin{array}{l}=({\stackrel{¯}{x}}_1−{\stackrel{¯}{x}}_2)Â\pm t_{\mathrm{Î\pm }/2}×\sqrt{\frac{{s_1}^2}{n_1}+\frac{{s_2}^2}{n_2}}\\ =(2.36−3.6)Â\pm 1.895×\sqrt{\frac{0.733}{5}+\frac{0.42}{4}}\\ =(−1.24)Â\pm (1.895×0.502)\\ =−1.24Â\pm 0.95\end{array}$

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

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 ${\mathrm{μ}}_2>{\mathrm{μ}}_1$without specifying any value of the difference between the population means.

Therefore, the confidence interval provides more information ${\mathrm{μ}}_1−{\mathrm{μ}}_2$.