Question

Reserve Problems Chapter 11 Section 2 Problem 2 The peanut crop was harvested from five fields of various area. The following data are the mass of the crop from each field y (in kilograms) and the field area x (in hectares). y 7320 15590 1368019920 12980 x 2.33 4.09 3.35 4.19 2.32 Round your intermediate answers to four decimal places (e.g. 98.7654). (a) Fit the simple linear regression model using the method of least squares. Find the estimate of o?. Round your answer to the nearest integer (e.g. 9876). (b) What change in the mean mass is expected when the field area changes by 1 hectare? Round your answer to the nearest integer (e.g. 9876). Ñ , (c) Calculate the fitted value of y corresponding to x = 4.19. Find the corresponding residual. Round your answer to the nearest integer (e.g. 9876). ỹ= Round your answer to the nearest integer (e.g. 9876). e = (d) Estimate the mean mass of the crop harvested from 4.0 hectares. Round your answer to the nearest integer (e.g. 9876). y =

Answer 1

a)

x	y	x2	xy	y2
2.33	7320	5.4289	17055.6	53582400
4.09	15590	16.7281	63763.1	243048100
3.35	13680	11.2225	45828	187142400
4.19	19920	17.5561	83464.8	396806400
2.32	12980	5.3824	30113.6	168480400
$\tiny \sum$= 16.28	$\tiny \sum$ = 69490	$\tiny \sum$ = 56.318	$\tiny \sum$ = 240225.1	$\tiny \sum$ = 1049059700

estimate of $\tiny \sigma^2$ = $\tiny \frac{1}{N-2}\sum (y-\bar{y})^2$

where N = 5, $\tiny \sigma^2$ = 27762560

b)

Using

  $\tiny \hat{\beta}_1 = \frac{\sum (x-\bar{x})(y-\bar{y})}{\sum (x-\bar{x})^2}$

putting the values in above expression,

$\tiny \hat{\beta}_1$ = 4219

c)

$\tiny \hat{\beta_0} = \bar{y}-\hat{\beta_1}\bar{x}$ = 160.936

$\hat{y}=\hat{\beta _0}+\hat{\beta _1}*x$, using this equation we get,

$\tiny \hat{y}= 160.936 +4219*4.19$ = 17839

For residual,

$\hat{e}_i=y_i-\hat{y_i}$

= 19920 - 17839

= 2081

d) Using the same equation as used in above in question,

$\tiny \hat{y}=160.936+4219*4$

=17037

Hope this helps!