In: Statistics and Probability
A fast food restaurant chain claims that lunch will be served within nine minutes of placing your order, or it is free. A sample of eight customers revealed the following waiting times in minutes:
10.1 | 9.3 | 9.2 | 10.2 | 9.3 | 9.6 | 9.4 | 8.8 |
At the 0.01 significance level, can we conclude that the waiting time for lunch exceeds nine minutes? Estimate the p-value.
a. State the null hypothesis and the alternate hypothesis.
H0: μ ≤
H1: μ >
b. State the decision rule for 0.01 significance level. (Round the final answer to 3 decimal places.)
Reject H0 if t > .
c. Compute the value of the test statistic. (Negative answer should be indicated by a minus sign. Round the final answer to 2 decimal places.)
Value of the test statistic
d. At the 0.01 significance level, can we conclude that the waiting time for lunch exceeds nine minutes?
(Click to select) Do not reject Reject H0. There is (Click to select) not enough enough evidence to conclude that the waiting time is longer than 9 minutes.
e. Estimate the p-value.
p-value is .
Values ( X ) | ||
10.1 | 0.3752 | |
9.3 | 0.0352 | |
9.2 | 0.0827 | |
10.2 | 0.5077 | |
9.3 | 0.0352 | |
9.6 | 0.0127 | |
9.4 | 0.0077 | |
8.8 | 0.4727 | |
Total | 75.9 | 1.5291 |
Mean
Standard deviation
To Test :-
H0: μ ≤ 9
H1: μ > 9
Test Statistic :-
t = 2.9501 2.95
Test Criteria :-
Reject null hypothesis if
Result :- Fail to reject null hypothesis
Conclusion :- There is insufficient evidence to conclude that the waiting time is longer than 9 minutes.
P value = P ( t > 2.950 ) = 0.0107