Consider a scenario where we analyze the pricing behavior of two common shares (
ID: 3022659 • Letter: C
Question
Consider a scenario where we analyze the pricing behavior of two common shares (one in XYZ corporation and one in MIM corporation). We have collected daily price changes for both stocks for the last 100 trading days. The average daily price change in the stock of XYZ corporation during that time period is $0.06 with the population standard deviation of $0.10. The stock of MIM corporation has the sample average daily price change of $0.03 and the population standard deviation of $0.09.
Please test the hypothesis that the average daily price change in XYZ corporation exceeds that of MIM corporation by $0.02, i.e. mean(XYZ) – mean (MIM) = 0.02
Please conduct a 90% confidence test to investigate the hypothesis. PLEASE ASSUME THAT THE TWO standard deviations listed here are the POPULATION standard deviations.
Question 4 options:
A) Fail to reject the null hypothesis.
B) Reject the null hypothesis
C) None of the above
What was the value of the test statistic you computed in your test?
A) 0.743
B) -0.881
C) 0.918
D) 1.297
E) None of the above
EXCEL USE IS OKAY. JUST TELL ME THE COMMANDS. PLEASE SHOW YOUR WORK!!!!!!
A) Fail to reject the null hypothesis.
B) Reject the null hypothesis
C) None of the above
Explanation / Answer
Formulating the null and alternative hypotheses,
Ho: u1 - u2 <= 0.02
Ha: u1 - u2 > 0.02
At level of significance = 0.1
As we can see, this is a right tailed test.
Calculating the means of each group,
X1 = 0.06
X2 = 0.03
Calculating the standard deviations of each group,
s1 = 0.1
s2 = 0.09
Thus, the standard error of their difference is, by using sD = sqrt(s1^2/n1 + s2^2/n2):
n1 = sample size of group 1 = 100
n2 = sample size of group 2 = 100
Also, sD = 0.013453624
Thus, the z statistic will be
z = [X1 - X2 - uD]/sD = 0.743294146
where uD = hypothesized difference = 0.02
Now, the critical value for z is
zcrit = 1.281551566
[This can be produced using =NORMSINV(0.10)]
As z < 1.28155, WE FAIL TO REJECT THE NULL HYPOTHESIS.
Hence,
OPTION A: Fail to reject the null hypothesis. [ANSWER]
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As above,
OPTION A: z = 0.743 [ANSWER]
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