Chapter 15, Problem 12P

In the Chapter Preview we discussed a study by Judge and Cable (2010) demonstrating a negative relationship between weight and income for a group of women professionals. The following are data similar to those obtained in the study. To simplify the weight variable, the women are classified into five categories that measure actual weight relative to height, from = thinnest to 5 = heaviest. Income figures are annual income (in thousands), rounded to the nearest $1,000.

a. Calculate the Pearson correlation for these data.

b. Is the correlation statistically significant? Use a two-tailed test with $\alpha =.05$ .

Weight (X)	Income (Y)
1	115
1	78
4	53
3	63
5	37
2	84
5	41
3	51
1	94
5	44

To determine

1. The Pearson’s correlation for the given data.
2. If the correlation is statistically significant using a two tailed test with $\alpha =.05$ .

Answer to Problem 12P

Solution:

a.

Weight$(X)$	Income$(Y)$	$XY$	$X^2$	$Y^2$
1	115	115	1	13225
1	78	78	1	6084
4	53	212	16	2809
3	63	189	9	3969
5	37	185	25	1369
2	84	168	4	7056
5	41	205	25	1681
3	51	153	9	2601
1	94	94	1	8836
5	44	220	25	1936
${\displaystyle \sum _{}^{}{X}_{}}=$ 30	${\displaystyle \sum _{}^{}Y=}$ 660	${\displaystyle \sum _{}^{}XY=}$ 1619	${\displaystyle \sum _{}^{}{X}_{}^2}=$ 116	${\displaystyle \sum _{}^{}{Y}^2=}$ 49566

Therefore, the Pearson’s correlation between the weight$(X)$and the income$(Y)$is $-0.9135$, which strong negative correlation is.

b. Since the test statistic is beyond the left tail of the critical value $-2.306$ , therefore we reject H₀. That is, the sample of $n=10$ pairs of observation indicates that the correlation is statistically significant.

Explanation of Solution

Given:

The Given data:

Weight$(X)$	Income$(Y)$
1	115
1	78
4	53
3	63
5	37
2	84
5	41
3	51
1	94
5	44

Formula Used:

  $r=\frac{N{\displaystyle \sum _{}^{}{X}_{}{Y}_{}-\left({\displaystyle \sum _{}^{}X}\right)}\left({\displaystyle \sum _{}^{}Y}\right)}{\sqrt{\left[N{\displaystyle \sum _{}^{}{X}_{}^2-{\left({\displaystyle \sum _{}^{}X}\right)}^2}\right]}\left[N{\displaystyle \sum _{}^{}{Y}_{}^2-{\left({\displaystyle \sum _{}^{}Y}\right)}^2}\right]}$

Calculations:

a. Now to compute the Pearson’s correlation between the weight $(X)$ and the income $(Y)$ ,we will use the below formula of correlation:

  $r=\frac{N{\displaystyle \sum _{}^{}{X}_{}{Y}_{}-\left({\displaystyle \sum _{}^{}X}\right)}\left({\displaystyle \sum _{}^{}Y}\right)}{\sqrt{\left[N{\displaystyle \sum _{}^{}{X}_{}^2-{\left({\displaystyle \sum _{}^{}X}\right)}^2}\right]}\left[N{\displaystyle \sum _{}^{}{Y}_{}^2-{\left({\displaystyle \sum _{}^{}Y}\right)}^2}\right]}$

Where:

  $N$ = Number of pairs of observations.

  ${\displaystyle \sum _{}^{}X}=\text{ Sum of X observations}$

  ${\displaystyle \sum _{}^{}Y=\text{Sum of Y observations}}$

  ${\displaystyle \sum _{}^{}XY}=\text{Sum of the products of paired observations }$

  ${\displaystyle \sum _{}^{}{X}_{}^2}=\text{Sum of squared X observations}$

  ${\displaystyle \sum _{}^{}{Y}_{}^2}=\text{Sum of squared Y observations}$

Weight$(X)$	Income$(Y)$	$XY$	$X^2$	$Y^2$
1	115	115	1	13225
1	78	78	1	6084
4	53	212	16	2809
3	63	189	9	3969
5	37	185	25	1369
2	84	168	4	7056
5	41	205	25	1681
3	51	153	9	2601
1	94	94	1	8836
5	44	220	25	1936
${\displaystyle \sum _{}^{}{X}_{}}=$ 30	${\displaystyle \sum _{}^{}Y=}$ 660	${\displaystyle \sum _{}^{}XY=}$ 1619	${\displaystyle \sum _{}^{}{X}_{}^2}=$ 116	${\displaystyle \sum _{}^{}{Y}^2=}$ 49566

  $\begin{array}{c}r=\frac{10\times 1619-30\times 660}{\sqrt{\left[10\times 116-{30}^2\right]\left[10\times 49566-{660}^2\right]}}\\ =\frac{16190-19800}{\sqrt{(1160-900)(495660-435600)}}\\ =\frac{-3610}{\sqrt{260\times 60060}}\\ =\frac{-3610}{3951.658}\\ =-0.9135\end{array}$

Therefore, the Pearson’s correlation between the weight$(X)$and the income$(Y)$is$-0.9135$, which is strong negative correlation.

b. Now to find if the correlation is statistically significant, we will follow the 4-step procedure of hypothesis testing as:

Step 1: The first step is to set up the hypothesis and determine the significance level

The null and alternative hypotheses are given below:

  $H_0:\rho =0$ i.e., the observed sample correlation coefficient is not significant of any correlation in the population.

  $H_a:\rho \ne 0$ i.e., the observed sample correlation coefficient is significant of correlation in the population.

Also the significance level is $\alpha =.05$ .

Step 2: The second step is to set the criteria for a decision.

Here $n=10$ and the two tailed critical value for $n-2=10-2=8df$ and $.05\text{ significance level is}$

  $t_{0.05,8}=\pm 2.306$

The Critical $t-value$

  $=\pm 2.306$ and the criteria for decision is, if the test statistic is less than the critical value then we fail to reject the null hypothesis else we reject the null hypothesis.

Step 3: The third step is to compute the test statistic. Now we have to calculate the test statistic. The test statistic is given below:

  $\begin{array}{c}t=\frac{r\sqrt{(n-2)}}{\sqrt{(1-r^2)}}\\ =\frac{-0.9135\sqrt{(10-2)}}{\sqrt{(1-{(-0.91354)}^2)}}\\ =\frac{-0.9135\times 2.8284}{\sqrt{1-0.8345}}\\ =\frac{-2.5838}{0.4068}\\ =-6.35\left(\text{rounded to two decimals}\right)\end{array}$

Step 4: The fourth step is to make a decision.

Since the test statistic is beyond the left tail of the critical value $-2.306$ , therefore we reject H₀. That is, the sample of $n=10$ pairs of observation indicates that the correlation is statistically significant.

Conclusion:

1. The correlation between the weight (X) and Income (Y) is $-0.9135$ .
2. Using a two-tailed test with $\alpha =.05$, we clearly see that the correlation is statistically significant.