Answer to Question #285261 in Statistics and Probability for Resia

Question

the following are the ages of husbands and wives for six couples. husbands age: (43, 57, 28, 19, 35, 39) wifes age: (37, 51, 32, 20, 33, 38)

Answer 1

a. We expect the ages of husbands and wives to be positively related.

b.

By looking at the scatter diagram, we expect the correlation coefficient between these two variables to be close to $1.$

c.

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c} & X & Y & XY & X^2 & Y^2 \\ \hline & 43 & 37 & 1591 & 1849 & 1369 \\ \hdashline & 57 & 51 & 2907 & 3249 & 2601 \\ \hdashline & 28 & 32 & 896 & 784 & 1024 \\ \hdashline & 19 & 20 & 380 & 361 & 400 \\ \hdashline & 35 & 33 & 1155 & 1225 & 1089 \\ \hdashline & 39 & 38 & 1482 & 1521 & 1444 \\ \hdashline Sum= & 221 & 211 & 8411 & 8989 & 7927 \\ \hdashline \end{array}$$

$$\bar{X}=\dfrac{1}{n}\displaystyle\sum_{i=1}^nX_i=\dfrac{221}{6}=36.833333$$

$$\bar{Y}=\dfrac{1}{n}\displaystyle\sum_{i=1}^nY_i=\dfrac{211}{6}=35.166667$$

$$SS_{XX}=\displaystyle\sum_{i=1}^nX_i^2-\dfrac{1}{n}(\displaystyle\sum_{i=1}^nX_i)^2=8989-\dfrac{221^2}{6}$$

$$=848.833333$$

$$SS_{YY}=\displaystyle\sum_{i=1}^nY_i^2-\dfrac{1}{n}(\displaystyle\sum_{i=1}^nY_i)^2=7927-\dfrac{211^2}{6}$$

$$=506.833333$$

$$SS_{XY}=\displaystyle\sum_{i=1}^nX_iY_i-\dfrac{1}{n}(\displaystyle\sum_{i=1}^nX_i)(\displaystyle\sum_{i=1}^nY_i)$$

$$=8411-\dfrac{221(211)}{6}=639.666667$$

Correlation coefficient:

$$r=\dfrac{SS_{XY}}{\sqrt{SS_{XX}}\sqrt{SS_{YY}}}$$

$$=\dfrac{639.666667}{\sqrt{848.833333}\sqrt{506.833333}}\approx 0.974474$$

The value of $r$ is consistent with what we expected in parts a and b.

$r=0.974474,$ strong positive correlation.

d. The following null and alternative hypotheses need to be tested:

$H_0:\rho=0$

$H_1:\rho\not=0$

where $\rho$ corresponds to the population correlation.

The sample size is $n = 6,$ so then the number of degrees of freedom is $df = n-2 = 6 - 2 = 4.$

The corresponding critical correlation value $r_c$ for a significance level of $\alpha = 0.05,$ for a two-tailed test is:

$$t=r\sqrt{\dfrac{n-2}{1-r^2}}$$

$$=0.974474\sqrt{\dfrac{6-2}{1-(0.974474)^2}}$$

The p-value for two-tailed, $df=4$ degrees of freedom, $t=8.681268,$ is computed as follows:

$$p=P(|t|>8.681286)$$

$$=0.000969$$

Since we have that $p = 0.000969 < 0.05=\alpha,$ it is concluded that the null hypothesis $H_0$ is rejected.

Therefore, based on the sample correlation provided, it is concluded that there is enough evidence to claim that the population correlation $\rho$ is different than $0,$ at the $\alpha = 0.05$ significance level.