Question

The management of JAL Airlines wants to determine if there is a linear relationship between advertising expenditures
and the number of passengers who choose to fly JAL.

The following data is collected over the past 15 months of performance by JAL Airlines.
Note that ADEXP represents Advertising Expenditures in $1,000s, and Passengers is the
number of passengers in 1,000s).

ADEXP	PASSENGERS
100	15
120	17
80	13
170	23
100	16
150	21
100	14
140	20
190	24
100	17
110	16
130	18
160	23
100	15
120	16

a)    Find the regression equation relating Passengers to ADEXP.

b)   Calculate the correlation coefficient, accurate to three decimal places  

c) Test the significance of the correlation coefficient with a 0.05 level of significance.
Include definition of hypotheses, test statistic, p-value or critical value, decision about Ho, and conclusion.

d)  What is the best estimate for number of passengers when $125,000 is spent on ads?

Answer 1

Suppose, random variables X and Y denote expenditure (in thousand dollars) in advertisements and number of passengers who choose to fly JAL Airlines respectively.

(a)

Regression equation of y on x is given by

$\tiny y-\bar y=b_{yx}\left(x-\bar x\right)$

Here,

Number of pairs of observation

n = 15

$\tiny \bar y=\frac{1}{n}\sum_{i=1}^{n}y_i=\frac{1}{15}\sum_{i=1}^{15}y_i=17.86667$

1 1 1 15 it = 2 = 124.6667 Tl i=1 15 i=1

$\tiny b_{yx}=\frac{Cov(X,Y)}{Var(X)}=0.1081317$

$\tiny \therefore y-17.86667=0.1081317\left(x-124.6667\right)$

$\tiny \Rightarrow y-17.86667=0.1081317x-0.1081317*124.6667$

$\tiny \Rightarrow y=0.1081317x+17.86667-13.48042$

$\tiny \therefore y=0.1081317x+4.38625$

(b)

Correlation coefficient is given by

$\tiny r_{x,y}=\frac{Cov(X,Y)}{\sqrt {Var(X)*Var(Y)}}=0.9683784$

(c)

We have to test for null hypothesis $\tiny H_0:\rho_{xy}=0$

against the alternative hypothesis $\tiny H_1:\rho_{xy}\neq0$

Our test statistic is given by

$\tiny t=\frac{r_{xy}\sqrt{n-2}}{\sqrt{1-r_{xy}^2}}$

Here,

Sample size

n = 15

Sample correlation coefficient is given by

$\tiny r_{x,y}=\frac{Cov(X,Y)}{\sqrt {Var(X)*Var(Y)}}=0.9683784$

$\tiny \therefore t_{calculated}=\frac{0.9683784\sqrt{13}}{\sqrt{1-0.9683784^2}}=13.99492$

Degrees of freedom $\tiny v=n-2=13$

$\tiny \text{p-value}=P\left(t_{13}\leq-13.99492\right)+P\left(t_{13}\geq13.99492\right)=0.000000003238011$

[Using R-code 'pt(-13.99492,13)+1-pt(13.99492,13)']

Level of significance $\tiny \alpha=0.05$

We reject our null hypothesis if $\tiny \text{p-value}<\alpha$

Here, we observe that $\tiny \text{p-value}=0.000000003238011\nless0.05=\alpha$

So, we reject our null hypothesis.

Hence, based on the given data we can conclude that there is significant evidence that there is a linear relationship between advertising expenditures and the number of passengers who choose to fly JAL.

(d)

For $\tiny x=125$ , predicted number of passengers is given by

$\tiny \hat y=0.1081317*125+4.38625=17.90271$

Hence, best estimate for number of passengers is 18.