91影视

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:

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

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

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

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

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

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

Therefore, the pooled estimate of variance is 0.6

Null Hypothesis,${\text{H}}_{\text{0}}{\text{:渭}}_{\text{1}}鈮�{\text{渭}}_{\text{2}}$and

Alternate Hypothesis,${\text{H}}_{\text{a}}{\text{:渭}}_{\text{1}}{\text{\&濒迟;渭}}_{\text{2}}$

The level of significance is $0.10$.

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

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 $-1.415$.

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

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

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

The 90% confidence interval for the difference in means

$\begin{array}{l}=({\stackrel{炉}{x}}_1鈭�{\stackrel{炉}{x}}_2)卤t_{伪/2}脳\sqrt{\frac{{s_1}^2}{n_1}+\frac{{s_2}^2}{n_2}}\\ =(2.36鈭�3.6)卤1.895脳\sqrt{\frac{0.733}{5}+\frac{0.42}{4}}\\ =(鈭�1.24)卤(1.895脳0.502)\\ =鈭�1.24卤0.95\end{array}$

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

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

Therefore, the confidence interval provides more information ${渭}_1鈭�{渭}_2$.