You manage a staff of 10 sales professionals for a distributor of IT products an

Question

You manage a staff of 10 sales professionals for a distributor of IT products and services. Over the past 6 months you have been tracking the number of client visits each sales person makes per week and have created a summary measure for average number of visits per week. Paired to this you have tallied sales in the first six months of the year for each sales person. As you look at the numbers, you postulate that there might be a relationship between the two variables and you formally want to investigate the possibility. The Data is listed below: SalesPerson Client Visits Per week YTD Sales Caleb 7.85 46.54 Josiah 14.87 67.21 Eileen 10.68 70.62 Chanda 10.1 51.03 Rodderick 13.83 74.5 Dimiya 12.73 60.79 Leena 6.84 57.08 Addison 7.72 38.51 Shah 4.06 42.02 John 1.71 23.07 Q7 – Can you reject the hypothesis that the slope for your model is = 0 at the 5% significance level? If significantly different what are the bounds of a 95% confidence interval for the slope? Q8 – Suppose John decided to see more clients in hope of improving his sales. He has been averaging 10 client visits per week, what would his predicted sales be? Q9 – What are the bounds of a 95% confidence interval for the predicted number of sales in Q8? Q10 – Suppose ALL your sales force averaged 10 sales calls per week, what are the bounds of a 95% confidence interval for the average predicted number of sales in Q8?

Explanation / Answer

Solution:

Q7:

We shall use the open source software R to solve this , please see the complete R snippet along with the results as

# read the data into R dataframe
data.df<- read.csv("C:\Users\586645\Downloads\Chegg\ytd.csv",header=TRUE)
str(data.df)

# perform anova analysis
a<- lm(YTD.Sales~ Client.Visits.Per.week,data=data.df)

#summarise the results
summary(a)

The results are

Call:
lm(formula = YTD.Sales ~ Client.Visits.Per.week, data = data.df)

Residuals:
Min 1Q Median 3Q Max
-10.255 -5.532 -3.619 5.459 12.044

Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 23.1761 6.4999 3.566 0.007340 **
Client.Visits.Per.week 3.3146 0.6573 5.043 0.000998 ***

as the p value is less than 0.05 , hence we can safely reject the null hypothesis for Q7 and conclude that the slope of the model is not 0

Q8:
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 8.337 on 8 degrees of freedom
Multiple R-squared: 0.7607,   Adjusted R-squared: 0.7308
F-statistic: 25.43 on 1 and 8 DF, p-value: 0.000998

The equation is

Y = 23.17 + 3.31*X

so put 10 as x , we get Y = 56.27

> x<- data.frame(Client.Visits.Per.week=10)
> predict(a,x)
1
56.32235
> predict(a,x,interval = "predict")
fit lwr upr
1 56.32235 36.10593 76.53877

Q10:

data.df$Client.Visits.Per.week<- 10

# perform anova analysis
a<- lm(YTD.Sales~ Client.Visits.Per.week,data=data.df)

#summarise the results
summary(a)

x<- data.frame(Client.Visits.Per.week=10)

predict(a,x,interval = "predict")

The results are

Call:
lm(formula = YTD.Sales ~ Client.Visits.Per.week, data = data.df)

Residuals:
Min 1Q Median 3Q Max
-30.067 -9.987 0.918 12.468 21.363

Coefficients: (1 not defined because of singularities)
Estimate Std. Error t value Pr(>|t|)
(Intercept) 53.137 5.081 10.46 2.46e-06 ***
Client.Visits.Per.week NA NA NA NA
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 16.07 on 9 degrees of freedom

> x<- data.frame(Client.Visits.Per.week=10)
> predict(a,x,interval = "predict")
fit lwr upr
1 53.137 15.01377 91.26023

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