Question

Merrill Lynch, a wealth management and financial services company, is considering recommending a new stock to its clients. Prior to recommending it, they want to be able to say that the stock has outperformed the market return of 7% in the last ten years. To be as detailed as possible, they use monthly return data for the past 3 years. The average return for the stock is 7.3% and the known population standard deviation for stocks is 3%. Note, use whole numbers for percent (example for 7.3% plug in 7.3 not .073)

Calculate the correct p-value for the test. Hint, you will need to use the appropriate table to find this and will require you to know alpha, and if it is a one or two-tailed test!

Answer 1

Claim: To test whether that the stock has outperformed the market return of 7% in the last ten years or not

Hypothesis: $Ho:\mu \leq 7$

$H1:\mu >7$

Right tailed test ( One tailed test )

Test statistics : $z =\frac{\bar{x}-\mu }{\sigma /\sqrt{n}}$

Where,

Sample mean $\bar x = 7.3$

Population SD $=\sigma=3$

Sample size = n = 3

$\alpha=0.05$

$z =\frac{7.3-7 }{3 /\sqrt{3}}=0.173$

Test statistics = z = 0.173

Pvalue : $P(Z>z)=P(Z>0.173)$

$=1-P(Z\leq0.173)$

$=1- 0.5687$ .........( From Z table)

  $=0.4313$

Decision Rule : $Pvalue>\alpha$ We Fail to Reject Ho

Conclusion : There is INSUFFICIENT evidence to conclude that the stock has outperformed the market return of 7% in the last ten years.

Answer 2

The claim is that the stock has outperformed the market return of lo = 7% in the last 10 years, that is, the test is right tailed. The hypotheses are as follows: The null hypothesis is H,:S7 VS the alternative hypothesis H:\>7. The monthly data is collected for 3 years, therefore, the sample size is n=3x12 = 36. The sample average return for the stock is t = 7.3%. The known population standard deviation is o = 3%. Since the population standard deviation is known, therefore Z-test is applicable. The Z-test statistics is calculated as follows: X - Ho on 7.3-7 3/36 0.3x6 3 z = 0.6 The P-value for the test is calculated as follows: P-value = P(Z >0.6) =1-P(Z 50.6) =1-0.7257 (From standard normal table) P-value = 0.2743