Question

The data for a random sample of 10 paired observations is shown below.

Pair	Sample from Population 1 (Observation 1)	Sample from Population 2 (Observation 2)
$\begin{array}{l}\mathbf{\text{1}}\\ \mathbf{\text{2}}\\ \mathbf{\text{3}}\\ \mathbf{\text{4}}\\ \mathbf{\text{5}}\\ \mathbf{\text{6}}\\ \mathbf{\text{7}}\\ \mathbf{\text{8}}\\ \mathbf{\text{9}}\\ \mathbf{\text{10}}\end{array}$	$\begin{array}{l}\mathbf{\text{19}}\\ \mathbf{\text{25}}\\ \mathbf{\text{31}}\\ \mathbf{\text{52}}\\ \mathbf{\text{49}}\\ \mathbf{\text{34}}\\ \mathbf{\text{59}}\\ \mathbf{\text{47}}\\ \mathbf{\text{17}}\\ \mathbf{\text{51}}\end{array}$	$\begin{array}{l}\mathbf{\text{24}}\\ \mathbf{\text{27}}\\ \mathbf{\text{36}}\\ \mathbf{\text{53}}\\ \mathbf{\text{55}}\\ \mathbf{\text{34}}\\ \mathbf{\text{66}}\\ \mathbf{\text{51}}\\ \mathbf{\text{20}}\\ \mathbf{\text{55}}\end{array}$

a. If you wish to test whether these data are sufficient to indicate that the mean for population 2 is larger than that for population 1, what are the appropriate null and alternative hypotheses? Define any symbols you use.

b. Conduct the test, part a, using$\mathbf{\text{α=.10}}$.

c. Find a $90\%$confidence interval for ${\mathit{\mu }}_{\mathbf{d}}$. Interpret this result.

d. What assumptions are necessary to ensure the validity of this analysis?

Short answer

A confidence interval is explained as the set of numbers seen in our sample for which we anticipate discovering the number that best represents the populace.

Step-by-step solution

Step-by-Step Solution Step 1: (a) State the null and alternate hypotheses

Let ${\mu }_1$and ${\mu }_2$be the means of populations 1 and 2, respectively, and ${\mu }_d$be the difference between the means of two populations.

Null Hypothesis $\left(H_0\right)$: There is no difference between the two population means.

$H_0:{\mu }_d\ge 0({\mu }_1-{\mu }_2\ge 0)$

Alternate Hypothesis $\left(H_0\right)$: The mean of population 2 is larger than population 1.

$H_a:{\mu }_d<0({\mu }_1-{\mu }_2<0)$

(b) Conduct the Test

$\begin{array}{l}\overline{d}=\frac{{\displaystyle \sum d}}{n}\\ =\frac{-37}{10}\\ =-3.7\end{array}$

$\begin{array}{l}{s}_d=\sqrt{\frac{{\displaystyle \sum d^2-\frac{{\left({\displaystyle \sum d}\right)}^2}{n}}}{n-1}}\\ =\sqrt{\frac{181-\frac{{(-37)}^2}{10}}{9}}\\ =\sqrt{\frac{44.1}{9}}\\ =2.2136\end{array}$

Here,$n=10$

So, the degree of freedom will be $=n-1=9$

$\alpha =0.10$

From the t-table, the critical value at $10\%$the level of significance with a degree of freedom 9 is $1.383$.

$\begin{array}{l}t=\frac{\overline{d}}{\frac{s_d}{\sqrt{n}}}\\ =\frac{-3.7}{\frac{2.2136}{\sqrt{10}}}\\ =-5.29\end{array}$

$\left|t\right|=5.29$

Since$5.29>1.383$ so null hypothesis will be rejected. Therefore, it can be concluded that the mean of population 2 is larger than population 1.

(c) Form the confidence interval

Here, $n=10$

So, the degree of freedom will be $=n-1=9$

$\alpha =0.10$

From the t-table, the critical value at $10\%$the level of significance with a degree of freedom $9$is role="math" localid="1652708493593" $1.383$.

The margin of error is,

$\begin{array}{l}ME=t_{\alpha /2}\left(\frac{s_d}{\sqrt{n}}\right)\\ =\left(1.833\right)\left(\frac{2.2136}{\sqrt{10}}\right)\\ =1.2831\end{array}$

The Confidence Interval is,

$\begin{array}{l}\mathrm{CI}=\overline{\mathrm{d}}\pm \mathrm{ME}\\ =(-3.7)\pm (1.2831)\\ =(-4.98,-2.42)\end{array}$

(d) State the assumption

• A random sample of differences is selected from the target population of differences.

• The population of differences has an approximately normal distribution.