In: Statistics and Probability
Can an answer, complete with workings please be provided for Problem 32E, Chapter 15 from the Business Statistics: Communicating with Numbers (2nd Edition) be made available?
The manager of a local Costo store is in the process of making hiring decisions for selling mobile phone contracts. She believes that the sale of mobile phone contracts depends crucially on the number of hours clocked by male and female employees. She collects the weekly data on last year’s sales of mobile phone contracts (Sale) along with work hours of male (Hours Males) and female (Hours Females) employees. A portion of the data is shown in the accompanying table.
a. |
Report the sample regression equation of the appropriate model. |
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b. |
At the 5% significance level, are the explanatory variables jointly significant? Are they individually significant? Use the p-value approach for the tests. |
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c. |
The manager would like to determine whether there is a difference in productivity of male and female employees. In other words, for the same work hours, whether the number of sales of mobile contracts varies between male and female employees. Conduct the appropriate test at the 5% level of significance. Provide the details.
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a)
Here i write R-code for given problem as:
s=c(59,65,62,61,65,63,55,61,63,63,59,63,64,61,62,61,62,62,62,63,65,59,62,59,61,61,65,62,56,63,64,59,63,61,64,62,56,63,63,63,64,59,61,64,64,65,62,63,65,61,61,64)
m=c(30,33,27,28,33,33,27,25,31,32,28,34,35,26,30,31,29,31,29,33,33,30,30,28,31,30,35,29,24,29,28,29,27,30,32,29,26,29,27,32,28,24,28,31,32,31,28,33,32,29,30,25)
f=c(32,36,34,34,37,35,29,28,34,35,29,34,34,35,33,33,34,32,34,32,36,29,35,33,32,35,32,34,31,36,34,33,36,35,35,34,32,35,37,34,38,33,35,34,36,35,36,34,36,36,34,35)
d=data.frame(s,m,f)
d
a=aov(s~f+m+f*m,d)
summary(a)
And the output is;
summary(a)
Df Sum Sq Mean Sq F value Pr(>F)
f 1 114.41 114.41 52.692 2.95e-09 ***
m 1 59.04 59.04 27.192 3.87e-06 ***
f:m 1 0.01 0.01 0.005 0.944
Residuals 48 104.23 2.17
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
b)
Here both p-values are less than 0.05, thus the explainitory
variable is jointly significant.
And the interaction effect is insignificant.
c)
H0: Their is no difference in productivity of male and female
employee.
H1: Their is difference in productivity of male and female
employee.
Here the R-code is as follows;
m=c(30,33,27,28,33,33,27,25,31,32,28,34,35,26,30,31,29,31,29,33,33,30,30,28,31,30,35,29,24,29,28,29,27,30,32,29,26,29,27,32,28,24,28,31,32,31,28,33,32,29,30,25)
f=c(32,36,34,34,37,35,29,28,34,35,29,34,34,35,33,33,34,32,34,32,36,29,35,33,32,35,32,34,31,36,34,33,36,35,35,34,32,35,37,34,38,33,35,34,36,35,36,34,36,36,34,35)
t.test(m,f)
And the output is,
Welch Two Sample t-test
data: m and f
t = -8.9488, df = 96.51, p-value = 2.59e-14
alternative hypothesis: true difference in means is not equal to
0
95 percent confidence interval:
-5.169160 -3.292378
sample estimates:
mean of x mean of y
29.69231 33.92308
Here, p-value is less than 0.05, thus we reject H0 at 5% l.o.s.