Question

Consider the following data for 15 subjects with two predictors. The dependent variable, MARK, is the total score for a subject on an examination. The first predictor, COMP, is the score for the subject on a so-called compulsory paper. The other predictor, CERTIF, is the score for the subject on a previous exam.

Student	MARK	COMP	CERTIF
1	476	111	68
2	457	92	46
3	540	90	50
4	551	107	59
5	575	98	50
6	698	150	66
7	545	118	54
8	574	110	51
9	645	117	59
10	556	94	97
11	634	130	57
12	637	118	51
13	390	91	44
14	562	118	61
15	560	109	66

a.Run a stepwise regression on the dataset

b.Does CERTIF add anything to predicting MARK, above and beyond that of COMP?

c. Write out the prediction equation

d. A statistician wishes to know the sample size needed in a multiple regression study. She has four predictors and can tolerate at most a .10 drop-off in predictive power. But she wants this to be the case with .95 probability. From previous related research, the estimated squared population multiple correlation is .62. How many subjects are needed?

Answer 1

a.Run a stepwise regression on the dataset

Load the data into Excel.

Go to Data>Megastat.

Select the option Correlation/Regression and go to Regression.

Select COMP and CERTIF as the independent variable(s), x.

Select MARK as the dependent variable, y.

Click OK.

The output will be as follows:

	R²	0.583
	Adjusted R²	0.514	n	15
	R	0.764	k	2
	Std. Error	54.463	Dep. Var.	MARK
ANOVA table
Source	SS	df	MS	F	p-value
Regression	49,791.1563	2	24,895.5782	8.39	.0052
Residual	35,594.8437	12	2,966.2370
Total	85,386.0000	14
Regression output					confidence interval
variables	coefficients	std. error	t (df=12)	p-value	95% lower	95% upper
Intercept	124.0641
COMP	3.5120	0.8975	3.913	.0021	1.5566	5.4674
CERTIF	0.8346	1.1346	0.736	.4761	-1.6375	3.3068

b.Does CERTIF add anything to predicting MARK, above and beyond that of COMP?

CERTIF add a value 0.8345 to predict MARK, which is beyond that of COMP.

c. Write out the prediction equation.

The prediction equation is:

MARK = 124.0641 + 3.5120*COMP + 0.8346*CERTIF

Or

The total score for a subject on an examination = 124.0641 + 3.5120*Score for the subject on a so-called compulsory paper + 0.8346*Score for the subject on a previous exam

d. A statistician wishes to know the sample size needed in a multiple regression study. She has four predictors and can tolerate at most a .10 drop-off in predictive power. But she wants this to be the case with .95 probability. From previous related research, the estimated squared population multiple correlation is .62. How many subjects are needed?

The formula for calculating sample size is:

n = p(1 - p)(z/E)²

We are given:

p = 0.62 and E = 0.1

The critical value at the 0.05 significance level is 1.96.

Subject needed = 0.62(1 - 0.62)(1.96/0.1)²

= 0.2356*(19.6)²

= 0.2356*384.16

= 90.5 = 91

Therefore, we need atleast 91 subjects.