I need the answers to numbers 610 REGRESSION ANALYSIS Jane M

I need the answers to numbers 6-10.

REGRESSION ANALYSIS Jane Morris is studying companies going public for the first time. She is particularly interested in the relationship between the size of the offering and tht price per share. She wants to now if there is a relationship betweent company and the initial price per share. She wonders if the size can p price per share. A sample of 15 companices that rececatly weant publie rever following information. redict the e revealed the COMPANY SIZE PRICE PER SHARE ISMILLIONS) 90 94.4 27.3 179.2 10.8 11.2 11.1 11.2 11.0 97.9 10 12 14 0.0 160.7 965 83.0 23,5 58,7 93.8 344 10.6 10.5 0.3 10.7 11.0 10.8

Solution

Regression Analysis

r²

0.217

n

15

r

0.466

k

1

Std. Error

0.281

Dep. Var.

y

ANOVA table

Source

SS

df

MS

F

p-value

Regression

0.2844

1  

0.2844

3.61

.0799

Residual

1.0249

13  

0.0788

Total

1.3093

14  

Regression output

confidence interval

variables

coefficients

std. error

   t (df=13)

p-value

95% lower

95% upper

Intercept

10.6665

0.1457

73.188

2.15E-18

10.3517

10.9814

x

0.0030

0.0016

1.899

.0799

-0.0004

0.0064

Predicted values for: y

95% Confidence Interval

95% Prediction Interval

x

Predicted

lower

upper

lower

upper

Leverage

150

11.1191

10.8311

11.4071

10.4476

11.7906

0.225

6).

Y=10.6665+0.003*X

7).

Slope =0.003

When size of the company increases by one unit(million), the price per share increases by 0.003.

8).

Y intercept =10.6665

When size of the company 0 unit(million), the price per share is 10.6665.

There is no practical importance to y intercept.

9).

When size =150,

Y=10.6665+0.003*150=11.1191

Predicted price per share =11.12

10).

Since the regression is not significant, F=3.61, P=0.0799, the model is not useful.

Regression Analysis
	r²	0.217	n	15
	r	0.466	k	1
	Std. Error	0.281	Dep. Var.	y
ANOVA table
Source	SS	df	MS	F	p-value
Regression	0.2844	1	0.2844	3.61	.0799
Residual	1.0249	13	0.0788
Total	1.3093	14
Regression output				confidence interval
variables	coefficients	std. error	t (df=13)	p-value	95% lower	95% upper
Intercept	10.6665	0.1457	73.188	2.15E-18	10.3517	10.9814
x	0.0030	0.0016	1.899	.0799	-0.0004	0.0064
Predicted values for: y
		95% Confidence Interval	95% Prediction Interval
x	Predicted	lower	upper	lower	upper	Leverage
150	11.1191	10.8311	11.4071	10.4476	11.7906	0.225