Let x be a random variable representing dividend yield of Au

Let x be a random variable representing dividend yield of Australian bank stocks. We may assume that x has a normal distribution with mc011-1.jpg = 2.7%. A random sample of 23 Australian bank stocks has a sample mean of mc011-2.jpg = 8.44%. For the entire Australian stock market, the mean dividend yield is mc011-3.jpg = 6.7%. Do these data indicate that the dividend yield of all Australian bank stocks is higher than 6.7%? Use mc011-4.jpg = 0.01. Find (or estimate) the P-value. a. 0.001 b. 0.500 c. 0.000 d. 0.999 e. 1.998

Solution

Formulating the null and alternative hypotheses,              
              
Ho:   u   <=   6.7  
Ha:    u   >   6.7  
              
As we can see, this is a    right   tailed test.      
              
Thus, getting the critical t,              
df = n - 1 =    22          
tcrit =    +   2.508324553      
              
Getting the test statistic, as              
              
X = sample mean =    8.44          
uo = hypothesized mean =    6.7          
n = sample size =    23          
s = standard deviation =    2.7          
              
Thus, t = (X - uo) * sqrt(n) / s =    3.090646982          
              
Also, the p value is              
              
p =    0.002670287 (or closest is 0.001)

Thus,

OPTION A: 0.001 [ANSWER]