Question

Financial   Energy   Utilities
10.76 12.89   11.98
15.05 13.91   5.86
17.21 6.43   13.46
5.03 11.23   9.82
19.59 18.93   3.95
8.21 20.73   3.44
10.45 9.60   7.11
6.75 17.40   15.70

A stock analyst wondered whether the mean rate of return of? financial, energy, and utility stocks differed over the past 5 years. He obtained a simple random sample of eight companies from each of the three sectors and obtained the? 5-year rates of return shown in the accompanying table? (in percent). Complete parts? (a) through? (d) below. State the null and alternative hypotheses. Normal probability plots indicate that the sample data come from normal populations. Are the requirements to use the? one-way ANOVA procedure? satisfied? Are the mean rates of return different at the ?=0.05 level of? significance? Use technology to find the? F-test statistic for this data set. Determine the? P-value and state the appropriate conclusion. Draw boxplots of the three sectors to support the results obtained in part (c).

Answer 1

Ans :

The requirement for anova is satisfied as data is continuous and follows normality.

Let and be the mean rate of return of financial, energy and utility stocks i.e mean of treatment 1, treament 2 and treatment 3

The null hypothesis for an anova always assumes the population means are equal.

Hence, the null hypothesis as:

i.e all treatment means are equal.

mean rate of return of financial, energy and utility stocks is not significantly different

and the alternative hypothesis is given by

i.e all treatment means are not equal.

mean rate of return of financial, energy, and utility stocks is significantly different

The given data is

	Financial	Energy	Utilities
	10.76	12.89	11.98
	15.05	13.91	5.86
	17.21	6.43	13.46
	5.03	11.23	9.82
	19.59	18.93	3.95
	8.21	20.73	3.44
	10.45	9.6	7.11
	6.75	17.4	15.7
Total	93.05	111.12	71.32
Mean	11.63125	13.89	8.915

Let Xij be the observations.

i = 1,2 ...r. r is total number of rows

j= 1,2,...t t is total number of treatment.

Here

r=8

t=3

N= Total observation

= r*t

=24

The mean of these treatment can be calculate as

= 93.05 / 8

= 11.63125

= 111.12 / 8

= 2.5

= 71.32 / 8

= 8.915

Now obtaining the grand mean ( )

= 11.47875

Now calculating Sum of Squares.

Total Sum of Squares (TSS)

=(10.76-11.47875)²+(15.05-11.47875)²+(17.21-11.47875)²​​​​​​​+....(7.11-11.47875)²​​​​​​​+(15.7-11.47875)²​​​​​​​

= 597.4595

Treatment Sum of Squares ( SST)

= 3*(11.63125-11.47875)²​​​​​​​+3*(13.89-11.47875)²​​​​​​​+3*(8.915-11.47875)²​​​​​​​

= 99.2815

Error Sum of Square ( SSE)

SSE = TSS - SST

= 597.4595 - 99.28158

= 498.1779

Obtaining Degree of Freedom

df( Treatment )= t-1

= 2

df(error)= N-t

= 21

df(Total)=N-1

= 23

Calculating Mean Square of Error

Treatment Mean Square ( MST)

= 99.28158 /

=49.6408

= 597.4595 / 23

= 25.9765

= 498.1779 / 21

= 23.7228

Now obtaining the test statistic( F- statistic)

= 2.092554

Obtaining the p-value for F statistic =  2.092554

and df =( t-1,N-1)= (2,23)

p-value = 0.148354

At 0.05 level of significance

Since p-value > 0.05

We failed to reject the null hypothesis.

Hence there is not sufficient evidence to conclude that the mean rate of return of financial, energy and utility stocks is significantly different

The excel output is obtained as follows

Data Analysis > Anova : Single Factor

The output is

Obtaining the boxplot

From the boxplot

a) The the box plot for financial stock is little right skewed.But there is no outlier.

b) The box plot for Energy stock is also little right skewed and there is on outlier.

c) Teh box plot for Utilities stock is quite symmetric and there is no outlier.

So all three boxplot indicates that the data is normally distributed.