Question

Question: Independent random samples selected from two normal populations produced the sample means and standard deviations shown below.

Sample 1	Sample 2
$$\begin{gathered} {\mathbf{\text{n}}}_{\mathbf{\text{1}}}\mathbf{\text{= 17}} \\ {\overline{\mathbf{\text{x}}}}_{\mathbf{\text{1}}}\mathbf{\text{= 5.4}} \\ {\mathbf{\text{s}}}_{\mathbf{\text{1}}}\mathbf{\text{= 3.4}} \end{gathered}$$	role="math" localid="1660287338175" $$\begin{gathered} {\mathbf{\text{n}}}_{\mathbf{\text{2}}}\mathbf{\text{= 12}} \\ {\overline{\mathbf{\text{x}}}}_2\mathbf{\text{=}}\mathbf{\text{7.9}} \\ {\mathbf{\text{s}}}_2\mathbf{\text{=}}\mathbf{\text{4.8}} \end{gathered}$$

a. Conduct the test${\mathit{H}}_{\mathbf{0}}\mathbf{:}\mathbf{(}{\mathbf{\mu }}_{\mathbf{1}}\mathbf{-}{\mathbf{\mu }}_{\mathbf{2}}\mathbf{)}\mathbf{>}\mathbf{}\mathbf{10}$against ${\mathit{H}}_{\mathit{a}}\mathbf{:}\mathbf{(}{\mathbf{\mu }}_{\mathbf{1}}\mathbf{-}{\mathbf{\mu }}_{\mathbf{2}}\mathbf{)}\ne \mathbf{10}$. Interpret the results.

b. Estimate${\mathbf{\mu }}_{\mathbf{1}}\mathbf{-}{\mathbf{\mu }}_{\mathbf{2}}$ using a 95% confidence interval

Short answer

Answer

A confidence interval denotes the likelihood that a population parameter will drop among two specified numbers.

Step-by-step solution

(a) Conduct the test 

For Sample 1

$$\begin{gathered} {\overline{x}}_1=5.4 \\ {n}_1=17 \\ {s}_1=3.4 \end{gathered}$$

For Sample 2

$$\begin{gathered} {\overline{x}}_2=7.9 \\ {n}_2=12 \\ {s}_2=4.8 \end{gathered}$$

The null hypothesis is$H_0:\left({\mu }_1-{\mu }_2\right)=10$, the alternate hypothesis is$H_0:\left({\mu }_1-{\mu }_2\right)\ne 10$

Let us assume the level of significance is 0.05.

The degree of freedom will be

$$\begin{gathered} =n_1+n_2-2 \\ =12+17-2 \\ =27 \end{gathered}$$

From the t-distribution table, the critical value at 0.05 thelevel of significance for 27 degrees of freedom is 2.052.

The pooled standard deviation is$s_p=\sqrt{\frac{\left(n_1-1\right){s_1}^2+\left(n_2-1\right){s_2}^2}{n_1+n_2-2}}$

$$\begin{gathered} =\sqrt{\frac{\left(17-1\right){\left(3.4\right)}^2+\left(12-1\right){\left(4.8\right)}^2}{17+12-2}} \\ =\sqrt{\frac{184.96+253.44}{27}} \\ =4.03 \\ t=\frac{\left(\overline{x_1}-{\overline{x}}_2\right)-\left({\mu }_1-{\mu }_2\right)}{s_p\sqrt{\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}} \\ =\frac{\left(5.4-7.9\right)-0}{4.03\sqrt{\left(\frac{1}{17}+\frac{1}{12}\right)}} \\ =\frac{-2.5}{1.52} \\ =-1.644 \end{gathered}$$

Since, $-1.644<2.052$, so the null hypothesis will not be rejected.

Therefore, we can conclude that $\left({\mu }_1-{\mu }_2\right)$.

(b) Find confidence interval. 

The 95% confidence interval for the difference in means

$$\begin{gathered} =\left({\overline{x}}_1-{\overline{x}}_2\right)\pm t_{\alpha /2}\times s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}} \\ =\left(5.4-7.9\right)\pm 2.052\times 4.03\sqrt{\frac{1}{17}+\frac{1}{12}} \\ =\left(-2.5\right)\pm \left(2.052\times 1.52\right) \\ =-2.5\pm 3.12 \end{gathered}$$

Thus, the confidence interval for the mean difference is $-5.62\text{to 0.62}$.