Please explain. There are 5 areas to answer. The green cells the answer must be
ID: 3175321 • Letter: P
Question
Please explain. There are 5 areas to answer.
The green cells the answer must be typed in four decimals.
Purple is drop down select. Answers can be either Reject H0 or Failure to reject H0
The blue cells the answer must be typed to either 2 or 3 decimals depending on whether the Z or t table is appropriate.
A sample of 11 discount brokers showed a sample mean price charged for a trade of 100 shares at $50 per share was $46.50. Assume a population standard deviation of $7.90. Test the hypothesis that the mean price charged for a trade of 100 shares at $50 per share is at most $42.00 at a 0.1000. (See exercise 44 on page 372 of your textbook for a similar problem.) For the hypothesis stated above... Question 1 what is the test statistic? Question 2 What is the decision? Question 3 what is the p-value? Fill in only one of the following statements. If the Z table is appropriate, p-value If the t table is appropriate, p-value sExplanation / Answer
Given that,
population mean(u)=42
standard deviation, =7.9
sample mean, x =46.5
number (n)=11
null, Ho: <=2
alternate, H1: >42
level of significance, = 0.1
from standard normal table,right tailed z /2 =1.282
since our test is right-tailed
reject Ho, if zo > 1.282
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 46.5-42/(7.9/sqrt(11)
zo = 1.88922
| zo | = 1.88922
critical value
the value of |z | at los 10% is 1.282
we got |zo| =1.88922 & | z | = 1.282
make decision
hence value of | zo | > | z | and here we reject Ho
p-value : right tail - ha : ( p > 1.88922 ) = 0.02943
hence value of p0.1 > 0.02943, here we reject Ho
ANSWERS
---------------
a. test statistic: 1.8892
b. reject Ho
c. p-value: 0.02943