Question

A stock analyst wondered whether the mean rate of return of​ financial, energy, and utility stocks...

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).

Financial   Energy   Utilities
10.73   12.89   11.98
15.12   13.96   5.86
17.21   6.33   13.46
5.07   11.23   9.82
19.50   18.93   3.95
8.16   20.73   3.44
10.38   9.60   7.11
6.75   17.40   15.70

​(a) State the null and alternative hypotheses. Choose the correct answer below.

B.

H0: μfinancial=μenergy=μutilities and H1: at least one of the means is different

Your answer is correct.

​(b) Normal probability plots indicate that the sample data come from normal populations. Are the requirements to use the​ one-way ANOVA procedure​ satisfied?

C.

​Yes, because there are k=3 simple random​ samples, one from each of k​ populations, the k samples are independent of each​ other, and the populations are normally distributed and have the same variance.

Your answer is correct.

​(c) Are the mean rates of return different at the α=0.05 level of​ significance?

Use technology to find the​ F-test statistic for this data set.

F0equals=2.08

​(Round to two decimal places as​ needed.)

Determine the​ P-value and state the appropriate conclusion below.

Since the​ P-value is

nothing ​,

there

is

is not

enough evidence to reject the null hypothesis.​ Thus, we

can

cannot

conclude that the mean rates of return are different at the α=0.05 level of significance.

​(Round to three decimal places as​ needed.)

Homework Answers

Answer #1

excel data analysis tool for one factor anova,steps are: write data>menu>data>data analysis>anova :one factor>enter required labels>ok, and following o/p Is obtained,

Anova: Single Factor
SUMMARY
Groups Count Sum Average Variance
financial 8 92.92 11.615 26.64471
energy 8 111.07 13.88375 24.02074
utilities 8 71.32 8.92 20.59
ANOVA
Source of Variation SS df MS F P-value F crit
Between Groups 99.00 2 49.5009 2.0841 0.149 5.78
Within Groups 498.78 21 23.752
Total 597.78 23

F stat = 2.08

p value=0.149

Since the​ P-value is 0.149 ,

there is not

enough evidence to reject the null hypothesis.​ Thus, we

cannot

conclude that the mean rates of return are different at the α=0.05 level of significance.

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