Question

1.   The Wall Street Journal’s Shareholder Scoreboard tracks the performance of 1000 major U.S. companies. The...

1.   The Wall Street Journal’s Shareholder Scoreboard tracks the performance of 1000 major U.S. companies. The performance of each company is rated based on the annual total return, including stock price changes and the reinvestment of dividends. Ratings are assigned by dividing all 1000 companies into five groups from a (top 20%), b (next 20%), to e (bottom 20%). Shown here are the one-year ratings for a sample of 60 of the largest companies.

A

B

C

D

E

5

8

15

20

12

a)   The study wants to test whether the largest companies differ in performance from the performance of the 1000 companies in the Shareholder Scoreboard. Clearly state the null and alternative hypotheses.
b)   Compute the test statistic.
Please copy your R code and the result and paste them here.

c)   At 5% significance level, compute the critical value for the test statistic and the p value for the test. Draw your conclusion.
Please copy your R code and the result and paste them here.

d)   Use the function chisq.test() in R to run the test directly to confirm your results above are correct.

Please copy your R code and the result and paste them here.

Homework Answers

Answer #1

Hypothesis -

H1 : At least one proportion is different.

Test statistic -

Where, Oi - Observed frequency.

Ei - Expected frequency.

Ei = N*Pi

Calculation -

Test statistic - 11.5

Critical value -

Significance level = = 0.05

Using Excel,   =CHIINV( , df ) ,    df = n - 1 = 5 - 1 = 4

   = CHIINV( 0.05 , 4 ) = 9.488

P-value -

Using Excel function,   =CHIDIST( test statistic, df )

p-value = CHIDIST( 11.5 , 4 ) = 0.021484

Decision about null hypothesis -

P-value = 0.021484 is less than significance level = 0.05

Reject null hypothesis.

Conclusion -

There is enough evidence to claim that largest companies differ in performance from the performance of the 1000 companies in the Shareholder Scoreboard.

Using R :

R code -

Observed = c( 5,8,15,20,12)      #observed frequencies
Expected = c(0.2,0.2,0.2,0.2,0.2) #expected proportions
chisq.test(x = Observed, p = Expected) #run the chi-squared test

Output -

Chi-squared test for given probabilities

data: Observed
X-squared = 11.5, df = 4,   p-value = 0.02148

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