Question

11. Consider the following set of n-6 observations of x and y given in Table 1 i. Graph the observations in a scatter plot on paper by hand by plotting x on the x-axis and y on the y-axis. Comment on the relationship. Using observations from Table 1, what are the sample means i and-? Using observations from Table 1, calculate the deviations and squared deviations from the sample mean for x: (x 11. iii. -x) and (x )for each i. iv. Using observations from Table 1, compute the following quantities (by hand or using a calculator) and comment on what they represent: a) 721-x)2 (xi-) i-1 (01-9) b)

Answer 1

11. (i) The plot is as below.

There seems to be a positive association between x and y, meaning that as x increases, y also tends to increase on average.

(ii) The sample means would be as below.

$\dpi{80} \bar{x} = \frac{4 + 6 + 10 + 9 + 5 + 3}{6} = \frac{37}{6} = 6.1667$.

$\dpi{80} \bar{y} = \frac{3 + 8 + 9 + 10 + 6 + 4}{6} = \frac{40}{6} = 6.6667$.

(iii) The table is as below. Some values of the mean deviation are adjusted for rounding errors (so that the sum of mean deviation is zero).

$\dpi{80} i$	$\dpi{80} x$	$\dpi{80} (x - \bar{x})$	$\dpi{80} (x - \bar{x})^2$
1	4	−2.1667	4.69458889
2	6	−0.1667	0.02778889
3	10	3.8334	14.69495556
4	9	2.8334	8.02815556
5	5	−1.1667	1.36118889
6	3	−3.1667	10.02798889

(iv) (a) $\dpi{80} \frac{1}{n} \sum^n_{i=1} (x_i - \bar{x})^2 = \frac{1}{6} (4.69458889+0.02778889+14.69495556+8.02815556$ $\dpi{80} +1.36118889+10.02798889)$

or $\dpi{80} \frac{1}{n} \sum^n_{i=1} (x_i - \bar{x})^2 = \frac{1}{6} (38.83466668)$ or $\dpi{80} \frac{1}{n} \sum^n_{i=1} (x_i - \bar{x})^2 = 6.472444447$ .

This value represents the variance in x-values, which is a measure of the spread-out of the x-values.

(b) The table is as below. The values of mean deviations of y is adjusted similarly.

$\dpi{80} i$	$\dpi{80} x$	$\dpi{80} (x - \bar{x})$	$\dpi{80} (y - \bar{y})$	$\dpi{80} (x - \bar{x})(y - \bar{y})$
1	4	−2.1667	−3.6666	7.94442222
2	6	−0.1667	1.3333	−0.22226111
3	10	3.8334	2.3333	8.94447222
4	9	2.8334	3.3333	9.44457222
5	5	−1.1667	−0.6667	0.77783889
6	3	−3.1667	−2.6666	8.44432222

$\dpi{80} \frac{1}{n} \sum^n_{i=1} (x_i - \bar{x})(y_i - \bar{y}) = \frac{1}{6}(7.94442222-0.22226111+8.94447222+9.44457222$$\dpi{80} +0.77783889+8.44432222)$

or $\dpi{80} \frac{1}{n} \sum^n_{i=1} (x_i - \bar{x})(y_i - \bar{y}) = \frac{1}{6}(35.33336666)$ or $\dpi{80} \frac{1}{n} \sum^n_{i=1} (x_i - \bar{x})(y_i - \bar{y}) = 5.888894443$ .

This is the covariance between x and y, which is a measure of the joint variability of x and y. As the covariance is positive, we may confirm our previous comment on the positive association to be correct.

(c) $\dpi{80} \widehat{\beta}_1 = \frac{\frac{1}{n} \sum^n_{i=1} (x_i - \bar{x})(y_i - \bar{y})}{\frac{1}{n} \sum^n_{i=1} (x_i - \bar{x})^2}$ , and putting the values, we have $\dpi{80} \widehat{\beta}_1 = \frac{5.888894443}{6.472444447}$ or $\dpi{80} \widehat{\beta}_1 = 0.909840863$ .

This is the slope coefficient of the regression, which measures by how much the y-value increase on average for a unit increase in the x value.

(d) $\dpi{80} \widehat{\beta}_0 = \bar{y} - \widehat{\beta}_1 \bar{x}$ , and putting the values, we have $\dpi{80} \widehat{\beta}_0 = 6.6667 - 0.909840863*6.1667$ or $\dpi{80} \widehat{\beta}_0 = 1.055984$ .

(v) The regression equation is as below.

$\dpi{80} \widehat{y}_i = \widehat{\beta}_0 + \widehat{\beta}_1 x_i$ or $\dpi{80} \widehat{y}_i = 1.055984 + 0.909840863 x_i$ .

This represents the relation between the average/expected value of y and x values. For x to be zero, the average value of y would be $\dpi{80} \widehat{y}_i = 1.055984 + 0.909840863 * 0 = 1.055984$ . Also, for a unit change in x, the y would increase by $\dpi{80} 0.909840863$ units on average.

To test the slope coefficient, we must find the standard error of the slope coefficient. This would be $\dpi{80} se(\widehat{\beta}_1) = \sqrt{\frac{\widehat{\sigma}^2}{\sum (x_i - \bar{x})^2}}$ , for $\dpi{80} \widehat{\sigma}^2 = \frac{\sum u^2}{n-2}$ (where u's are residuals, and n-2 is the degree of freedom) or $\dpi{80} \widehat{\sigma}^2 = \frac{\sum (y_i - \widehat{y}_i)^2}{n-2}$ .

Then, we would have $\dpi{80} \frac{\widehat{\beta}_1 - \beta_1}{se(\widehat{\beta}_1)} \sim t_{\alpha,n-2}$ follows the t-distribution with n-2 degree of freedom. To test for $\dpi{80} \beta_1 = 0$ , the null hypothesis would be $\dpi{80} H_0 : \beta_1 = 0$ and $\dpi{80} H_1 : \beta_1 \ne 0$ , and the t-statistic would be $\dpi{80} t = \frac{\widehat{\beta}_1 - 0}{se(\widehat{\beta}_1)} = \frac{\widehat{\beta}_1}{se(\widehat{\beta}_1)}$ , and if the $\dpi{80} |t| > t_{\alpha,n-2}$ , we may reject the null.

(vi) The graph is as below.

The intercept value is $\dpi{80} \widehat{\beta}_0 = 1.055984$ , which is where the regression line touches the y-axis, which is average y value when x=0. The slope is $\dpi{80} \widehat{\beta}_1 = 0.909840863$ , which is the increase in average value of y, for a unit increase in x.