Question

I wonder if there is enough evidence that the average departure delay is less than 3.073 hours?

Null hypothesis H0:

Alternative hypothesis Ha:

alpha = 5%

Average= -4.37, S= 5.88, n=30

what is the P value?

What is the Critical value?

How did your test statistic compare to your critical value?

Reject or Fail to reject

Conclusion:

Answer 1

Solution :

Given that,

Population mean = = 3.073

Sample mean = = -4.37

Sample standard deviation = s = 5.88

Sample size = n = 30

Level of significance = = 0.05

This is a left (One) tailed test,

The null and alternative hypothesis is,  

Ho: 3.073

Ha: 3.073

The test statistics,

t = ( - )/ (s/)

= ( -4.37 - 3.073 ) / 5.88 / 30 )

= -6.933

Critical value of  the significance level is α = 0.05, and the critical value for a left-tailed test is

= -1.699

Since it is observed that t = -6.933 < = -1.699, it is then concluded that the null hypothesis is rejected.

P- Value = 0.000   

The p-value is p = 0 < 0.05, it is concluded that the null hypothesis is rejected.

Conclusion :

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the average departure delay is less than 3.073 hours, at the 0.05 significance level.