Question

Question 1. Online shopping statistics are routinely reported by www.shop.org. Of interest to many online retailers are gender-based differences in shopping preferences and behaviours. The summary data of monthly online expenditures for a sample of male and female online shoppers are shown in the following table:

	Male	Female
n	45	45
Mean	$352	$310
StDev s	$95	$80

Find the 90% confidence interval for the difference in mean monthly online expenditures between males and females.

  lower bound of confidence interval (use 2 decimal places in your answer).
  upper bound of confidence interval (use 2 decimal places in your answer).

Question 2

In the answer boxes below the table, provide the values for the table cells that have question marks.

	xbar	sx	ybar	sy	r	intercept of least sq. line	slope of least sq. line
a)	30	4	18	6	-0.2	??	??
b)	100	18	60	10	0.9	??	??
c)	??	0.8	50	15	??	-10	15
d)	??	??	18	4	-0.6	30	-2

Row a).

intercept
slope

Row b).

intercept
slope

Row c).

xbar
r

Row d).

xbar
s_x

Answer 1

ANs 1 ) using minitba>stat>basic stat>two sample t

we have

Two-Sample T-Test and CI

SE
Sample N Mean StDev Mean
1 45 352.0 95.0 14
2 45 310.0 80.0 12

Difference = μ (1) - μ (2)
Estimate for difference: 42.0
90% CI for difference: (11.2, 72.8)
T-Test of difference = 0 (vs ≠): T-Value = 2.27 P-Value = 0.026 DF = 88
Both use Pooled StDev = 87.8208

lower bound of confidence interval =11.20
  upper bound of confidence interval =72.8

Ans 2 )slope = r*sx/sy , intercept= y bar- slope*xbar

	xbar	sx	ybar	sy	r	intercept of least sq. line	slope of least sq. line
a)	30	4	18	6	-0.2	intercept = 18-(-0.3)*30= 27	slope = -0.2*6/4 = -0.3
b)	100	18	60	10	0.9	intercept = 60-(0.5)*100= 10	slope = 0.9*10/18 =0.5
c)	=(50-(-10))/15= 4	0.8	50	15	r= 15*0.8/15 =0.8	-10	15
d)	= (18-30)/-2= 6	=-0.6*4/-2 = 1.2	18	4	-0.6	30	-2