Question

You may use R or any other software to calcuate the sample variances. Suppose that the percentage changes in value per day on holdings of the stock of company ABC and DEF are normally distributed with unknown mean µabc, µdef and unknown variance σ2abc, σ2def. The following daily returns are observed over two business weeks (10 trading days): abc_pctchange <- c(2.82, 3.61, 5.66, -2.17, 2.40, -1.48, 3.55, 1.64, -0.16, 2.21)
def_pctchange <- c(1.83, 2.56, 3.55, -1.19, 0.40, -1.53, 2.44, 1.89, -0.54, 9.25) [the units are percentage points] 2.a) Produce a 95% confidence interval for the difference in mean return of the stock of company ABC and EDF, asuming that the variance are equal but unkown. (5 points) 2.b) Conduct a hypothesis test for the mean change in value stock ABC and DEF at a sgnificance level of 0.05, once again assuming that the variances are equal. State your hypothesis, calculate the p-value and state your conclusion. (5 points) 2.c)Did your confidence interval from a) contain the null hypothesis? Did your rejection region from c) contain the observed value of the test statistic? Are these two results contradictory? (5 points)a 2.d)Perform a F-test for the equality of variance based on the sample variances, can you reasonably assume that the variances are equal at a 0.05 significance level? (5 points)

Answer 1

abc_pctchange <- c(2.82, 3.61, 5.66, -2.17, 2.40, -1.48, 3.55, 1.64, -0.16, 2.21)
def_pctchange <- c(1.83, 2.56, 3.55, -1.19, 0.40, -1.53, 2.44, 1.89, -0.54, 9.25)

t.test(abc_pctchange,def_pctchange,var.equal = T)

        Two Sample t-test

data:  abc_pctchange and def_pctchange
t = -0.046474, df = 18, p-value = 0.9634
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 -2.679945  2.563945
sample estimates:
mean of x mean of y 
    1.808     1.866 

var.test(abc_pctchange,def_pctchange, alternative = "two.sided")

a)

95% confidence interval is (-2.679945 , 2.563945)

b)

Ho :

p-value = 0.9634

since p-value > alpha

we fail to reject the null hypothesis

c)

0 is present in confidence interval

hence we fail to reject the null hypothesis

d)

var.test(abc_pctchange,abc_pctchange, alternative = "two.sided")

F test to compare two variances

data:  abc_pctchange and def_pctchange
F = 0.61153, num df = 9, denom df = 9, p-value = 0.4752
alternative hypothesis: true ratio of variances is not equal to 1
95 percent confidence interval:
 0.1518963 2.4620306
sample estimates:
ratio of variances 
         0.6115336 

p-value = 0.4752 > alpha

hence we fail to reject the null hypothesis

we can reasonably assume that the variances are equal at a 0.05 significance level