Question

# Question 1. Online shopping statistics are routinely reported by www.shop.org. Of interest to many online retailers...

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
sx

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

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