Question

A statistical analyst for the Wall Street Journal randomly selected six companies and recorded both the price per share of stock on January 1, 2009 and on April 30, 2009. The results are presented below. Suppose the analyst wished to see if the average price per share of stock on April 30, 2009 is greater than the average price per share of stock on January 1, 2009 at α=.025.

Apr. 30, 2009	33	33	34	30	33	38
Jan. 1, 2009	21	25	30	33	23	27

For the hypothesis stated above, what is the decision (in terms of "April 30, 2009" minus "January 1, 2009")?

	a.	Reject H0 because P-value > α
	b.	None of the answers is correct
	c.	Reject H0 because the test statistic is to the right of the positive critical value
	d.	Fail to reject H0 because P-value > α
	e.	Fail to reject H0 because the test statistic is to the right of the positive critical value

Answer 1

Answer is:

OPTION C

The data:

Stock	Apr. 30, 2009	Jan. 1, 2009	Difference
1	33	21	12
2	33	25	8
3	34	30	4
4	30	33	-3
5	33	23	10
6	38	27	11

For the score differences we have, mean is Dˉ=7, the sample standard deviation is sD=5.6569, and the sample size is n=6. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: μD =0 Ha: μD >0 This corresponds to a Right-tailed test, for which a t-test for two paired samples be used. (2a) Critical Value Based on the information provided, the significance level is α=0.025, and the degree of freedom is n-1=6-1=5. Therefore the critical value for this Right-tailed test is tc​=2.5706. This can be found by either using excel or the t distribution table. (2b) Rejection Region The rejection region for this Right-tailed test is t>2.5706 (3)Test Statistics The t-statistic is computed as follows: (4) The p-value The p-value is the probability of obtaining sample results as extreme or more extreme than the sample results obtained, under the assumption that the null hypothesis is true. In this case, the p-value is 0.0145 (5) The Decision about the null hypothesis (a) Using traditional method Since it is observed that t=3.0311 > tc​=2.5706, it is then concluded that the null hypothesis is rejected. (b) Using p-value method Using the P-value approach: The p-value is p=0.0145, and since p=0.0145≤0.025, it is concluded that the null hypothesis is rejected. (6) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean μ1​ is greater than μ2, at the 0.025 significance level.