Question

To apply the two-sample t procedures, use Option 1 if you have technology that implements that method. Otherwise, use Option 2 (degrees of freedom the smaller of n1 - 1 and n2 – 1). Does the way the press depicts the economic future affect the stock market? To investigate this, researchers analyzed the longest article about the crisis from the front page of the Money section of USA Today from a randomly chosen weekday of each week between August 2007 and June 2009. Articles were rated as to how positive or negative they were about the economic future. For each week, the change in the Dow Jones Industrial Average (average DJIA value the week after the article appeared minus the average the week the article appeared) was computed. Positive values of the change indicate that the DJIA increased. Here are the changes in DJIA corresponding to very positive articles. (Data set may be found here.) -325 -200 -225 -75 -25 25 50 225 25 -225 -250 200 250 75 Here are the changes in DJIA values corresponding to very negative articles. 150 300 225 125 -175 -225 -375 -175 0 125 175 475 Is there good evidence that the DJIA performs differently after very positive articles than after very negative articles? (a) Find the sample means for very positive and very negative articles. (Round your answers to two decimal places.) positive articles x= negative articles ž=

Answer 1

Sr. No	Positive	P^2	Negative	N^2
1	-325	105625	150	22500
2	-200	40000	300	90000
3	-225	50625	225	50625
4	-75	5625	125	15625
5	-25	625	-175	30625
6	25	625	-225	50625
7	50	2500	-375	140625
8	225	50625	-175	30625
9	25	625	0	0
10	-225	50625	125	15625
11	-250	62500	175	30625
12	200	40000	475	225625
13	250	62500
14	75	5625
Total	-475	478125	625	703125
n	14		12
Mean	-33.929		52.083
SD	188.518		246.903

Mean = $1586279082626_blob.png$

SD =

Σχ2 - (Σα)2 VV a η -1

Positive mean = -33.929

Negative mean = 52.083

Here is a procedure to carry out the test

t-test for diff

$1586279013278_blob.png$ : There is no difference in performance based on articles. ( )

Mi = 12

$H_{1}$:There is difference in performance based on articles. ($\mu_{1}\neq\mu_{2}$ )

This is two-sided t-test for difference in population means. We use t-test because we do not have population SD. Aslso we assume that the population variances are same.

Pooled variance =

Vari(nı -1) + Varz(n2 - 1) ni + n2 - 2

= 47190.91

Test Stat =

(11 - 12) - (41 - H2) Pooled Var(1/n1 + 1/n2)

   = -1.006

p-value = 2P ( > T.S. )

tn+12-2

= 2P($t_{24}$ > 1.01) ...............using t-dist tables df = 12+14-2=24 and x = 1.01

= 2 * 0.1621

p - value =0.3242

We would reject the null hypothesis that have insufficient evidence if p-value < level of significance.