Answer:
d) At α=0.025, we fail to reject H0.
True. Since the p value is higher than the significance level 0.0275>0.025 we have enough evidence to FAIL to reject the null hypothesis.
Explanation:
1) Data given and notation
represent the sample mean
represent the sample standard deviation
sample size
represent the value that we want to test
represent the significance level for the hypothesis test.
t would represent the statistic (variable of interest)
represent the p value for the test
2) State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the mean is higher than 36 minutes :
Null hypothesis:
Alternative hypothesis:
Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:
(1)
t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
3) Calculate the statistic
We don't have info provided to calculate it but we can assume that is
4) P-value
First we need to calculate the degrees of freedom given by:
Since is a right tailed test the p value would be:
5) Conclusion
a) At α=0.05, we fail to reject H0.
False. On this case since the p value is lower than the significance level 0.0275<0.05 we have enough evidence to reject the null hypothesis not the opposite.
b) At α=0.02, we reject H0.
False. Since the p value is higher than the significance level because 0.0275>0.02 we FAIL to reject the null hypothesis not the opposite.
c) At α=0.03, we fail to reject H0.
False, since the p value is lower than the significance level because 0.0275<0.03 we have enough evidence to reject the null hypothesis not the opposite.
d) At α=0.025, we fail to reject H0.
True. Since the p value is higher than the significance level 0.0275>0.025 we have enough evidence to FAIL to reject the null hypothesis.