Question

The authors of an article found that the speed of a prey (twips/s) and the length of a prey (twips ✕ 100) are good predictors of the time (s) required to catch the prey. (A twip is a measure of distance used by programmers.) Data were collected in an experiment in which subjects were asked to "catch" an animal of prey moving across his or her computer screen by clicking on it with the mouse. The investigators varied the length of the prey and the speed with which the prey moved across the screen.

The following data are consistent with summary values and a graph given in the article. Each value represents the average catch time over all subjects. The order of the various speed-length combinations was randomized for each subject.

Prey Length	Prey Speed	Catch Time
7	20	1.11
6	20	1.21
5	20	1.24
4	20	1.41
3	20	1.49
3	40	1.40
4	40	1.35
6	40	1.30
7	40	1.28
7	80	1.41
6	60	1.37
5	80	1.39
7	100	1.43
6	100	1.42
7	120	1.71
5	80	1.49
3	80	1.40
6	100	1.51
3	120	1.91

(a) Fit a multiple regression model for predicting catch time using prey length and speed as predictors. (Use x₁ for prey length and x₂ for speed. Round your answers to three decimal places.)

=( ____) +(____  x₁₎ +( _____  x₂₎

(b) Predict the catch time for an animal of prey whose length is 6 and whose speed is 50. (Round your answer to three decimal places.)

(_____ s)

(C) Calculate the test statistic. (Round your answer to two decimal places.)

F = (_____)

(d) The authors of the article suggest that a simple linear regression model with the single predictor

x = length / speed

might be a better model for predicting catch time. Calculate the x values and use them to fit this linear regression model. (Round your answers to three decimal places.)

= (____) + (_____ x)

Answer 1

we will use excel to perform regression analysis

the output of regression analysis

SUMMARY OUTPUT
Regression Statistics
Multiple R	0.8582
R Square	0.7365
Adjusted R Square	0.7036
Standard Error	0.0958
Observations	19.0000
ANOVA
	df	SS	MS	F	Significance F
Regression	2.0000	0.4102	0.2051	22.3648	0.0000
Residual	16.0000	0.1467	0.0092
Total	18.0000	0.5569
	Coefficients	Standard Error	t Stat	P-value	Lower 95%	Upper 95%
Intercept	1.4342	0.0856	16.7447	0.0000	1.2526	1.6157
Prey Length	-0.0509	0.0150	-3.3927	0.0037	-0.0827	-0.0191
Prey Speed	0.0040	0.0006	6.2130	0.0000	0.0026	0.0053

a)

Catch Time=1.434-0.051*Prey Length+0.004*Prey Speed

b)

Predict the catch time for an animal of prey whose length is 6 and whose speed is 50. (Round your answer to three decimal places.)

Catch Time=1.434-0.051*Prey Length+0.004*Prey Speed

prey whose length is 6 and whose speed is 50

Catch Time=1.434-0.051*6+0.004*50

Catch Time = 1.328

(C) Calculate the test statistic. (Round your answer to two decimal places.)

F = 22.36

d)

single predictor

x = length / speed

create new variable

x
0.3500
0.3000
0.2500
0.2000
0.1500
0.0750
0.1000
0.1500
0.1750
0.0875
0.1000
0.0625
0.0700
0.0600
0.0583
0.0625
0.0375
0.0600
0.0250

output

SUMMARY OUTPUT
Regression Statistics
Multiple R	0.7169
R Square	0.5139
Adjusted R Square	0.4853
Standard Error	0.1262
Observations	19.0000
ANOVA
	df	SS	MS	F	Significance F
Regression	1.0000	0.2862	0.2862	17.9741	0.0006
Residual	17.0000	0.2707	0.0159
Total	18.0000	0.5569
	Coefficients	Standard Error	t Stat	P-value	Lower 95%	Upper 95%
Intercept	1.5831	0.0496	31.8906	0.0000	1.4783	1.6878
x	-1.3686	0.3228	-4.2396	0.0006	-2.0497	-0.6875

Catch Time = 1.583-1.369*x