Question

16) Serial correlation, also known as autocorrelation, describes the extent to which the result in one period of a time series is related to the result in the next period. A time series with high serial correlation is said to be very predictable from one period to the next. If the serial correlation is low (or near zero), the time series is considered to be much less predictable. For more information about serial correlation, see the book Ibbotson SBBI published by Morningstar.

An Internet advertising agency is studying the number of "hits" on a certain web site during an advertising campaign. It is hoped that as the campaign progresses, the number of hits on the web site will also increase in a predictable way from one day to the next. For 10 days of the campaign, the number of

hits × 10⁵

is shown.

Original Time Series

Day	1	2	3	4	5	6	7	8	9	10
Hits × 105	1.3	3.5	4.4	7.2	6.9	8.3	9.0	11.2	13.2	14.8

(a) To construct a serial correlation, we use data pairs

(x, y)

where x = original data and y = original data shifted ahead by one time period. Construct the data set

(x, y)

for serial correlation by filling in the following table.

x	1.3	3.5	4.4	7.2	6.9	8.3	9.0	11.2	13.2
y

(b) For the

(x, y)

data set of part (a), compute the equation of the sample least-squares line

ŷ = a + bx.

(Use 4 decimal places.)

a
b

If the number of hits was

9.2 (× 10⁵)

one day, what do you predict for the number of hits the next day? (Use 1 decimal place.)

(× 10⁵) hits

(c) Compute the sample correlation coefficient r and the coefficient of determination

r².

(Use 4 decimal places.)

r
r2

Test

ρ > 0

at the 1% level of significance. (Use 2 decimal places.)

t
critical t

Answer 1

(a)

data set (x, y) for serial correlation

x	1.3	3.5	4.4	7.2	6.9	8.3	9	11.2	13.2
y	3.5	4.4	7.2	6.9	8.3	9	11.2	13.2	14.8

(b)

the equation of the sample least-squares line,

= a + bx

n = 9

x	y	xy	x2
1.3	3.5	4.55	1.69
3.5	4.4	15.4	12.25
4.4	7.2	31.68	19.36
7.2	6.9	49.68	51.84
6.9	8.3	57.27	47.61
8.3	9	74.7	68.89
9	11.2	100.8	81
11.2	13.2	147.84	125.44
13.2	14.8	195.36	174.24
=65	=78.5	=677.28	=582.32

a= 1.6625

b=0.9775

If the number of hits was 9.2 (× 10⁵) one day, predict for the number of hits the next day? (Use 1 decimal place.) (× 10⁵) hits

the equation of the sample least-squares line,

= 1.6625 + 0.9775x

substitute x =9.2 in the above sample least-squares line

= 1.6625 + 0.9775 x 9.2 =10.6555 10.7

If the number of hits was 9.2 (× 10⁵) one day, predicted number of hits the next day 10.7(× 10⁵) hits

(c)

Compute the sample correlation coefficient r and the coefficient of determination r².

x	y	y2
1.3	3.5	12.25
3.5	4.4	19.36
4.4	7.2	51.84
7.2	6.9	47.61
6.9	8.3	68.89
8.3	9	81
9	11.2	125.44
11.2	13.2	174.24
13.2	14.8	219.04
		y2 =799.67

r = 0.9685

r² = 0.9685² = 0.9380

Null hypothesis : H_o : =0

Alternate Hypothesis : H_o : > 0

Right tailed test

Test statistic :

test statistic : t=10.3

Degrees of freedom : n-2 =9-2 =7

Level of significance : =0.01

For right tailed test ; Critical value of t at =0.01 for 7 degrees of freedom = 2.998

As value of the test statistic : 10.3 > Critical value of t : 2.998 ; Reject the null hypothesis

There is sufficient evidence to conclude that   > 0