Question

When 75 patients were randomly selected, the mean length of wait time was found to be 28 days and the standard deviation was 5.6 days. Because of waiting issues, wait times would slightly go up. What would be the steps to test the claim that wait times have a mean of LESS THAN 38 days if you used a 0.05 significance level?

Answer 1

Solution :

Given that,

Population mean = = 38

Sample mean = = 28

Sample standard deviation = s = 5.6

Sample size = n = 75

Level of significance = = 0.05

This is a left (One) tailed test,

The null and alternative hypothesis is,  

Ho: 38

Ha: 38

The test statistics,

t = ( - )/ (s/)

= ( 28 - 38 ) / ( 5.6 /75 )

= -15.465

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

= -1.666

Since it is observed that t = -15.465 = -1.666, it is then concluded that the null hypothesis is rejected.

P- Value = 0   

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 wait times have a mean of less than 38, at the 0.05 significance level.