A sample of 17 discount brokers showed a sample mean price charged for a trade o
ID: 3057516 • Letter: A
Question
A sample of 17 discount brokers showed a sample mean price charged for a trade of 100 shares at $50 per share was $35.80. Assume a population standard deviation of $5.20. Test the hypothesis that the mean price charged for a trade of 100 shares at $50 per share is at most $33 at =0.0100.
For the hypothesis stated above
1) What is the test statistic?
*** Answer must be typed to either 2 or 3 decimals depending on whether the Z or T table is appropriate*****
2) What is the decision? (Reject H0 or Fail to Reject H0 )
3) What is the p-value?
*** Answer must be typed to four decimals*****
If the Z Table is appropiate , p-vlue=_________
If the T Table is appopriate , _______ < p-value < ________
Explanation / Answer
sample mean is X¯=35.80 and the standard deviation = 5.20, and the sample size is n = 17.
(1) Null and Alternative Hypotheses
The following null and alternative hypotheses need to be tested:
Ho: = 33
Ha: < 33
This corresponds to a left-tailed test, for which a t-test for one mean.
(2) Rejection Region
Based on the information provided, the significance level is =0.01, and the critical value for a left-tailed test is tc=2.583.
The rejection region for this left-tailed test is R = {t: t < -2.583}
(3) Test Statistics
The t-statistic is computed as follows:
t = (X¯0) / ( /n)= (35.8033) / (5.20/17)=2.22
(4) Decision about the null hypothesis
Since it is observed that t=2.22 tc=2.583, it is then concluded that the null hypothesis is not rejected.
Using the P-value approach: The p-value is p = 0.9794, and since p = 0.9794 0.01, it is concluded that the null hypothesis is not rejected.
(5) Conclusion
It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean is at most 33, at the 0.01 significance level.