Question

At LIU, they wish to test whether the mean working hours/week of students in LIU Beirut is
higher than the mean working hours/week of students in Bekaa campus. A sample was taken
from each campus and the details were reported in the table below.

Beirut students (1) Bekaa students (2)

Population standard deviation σ1 = 6 σ2 =5
Sample Mean x̄1 = 35 x̄2 = 30
Sample size n1 = 40 n2 = 40
a. Can you conclude, at α = 0.1(10%), that students in Beirut have a higher mean working
hours/week than that of the students in Bekaa? (Use the 5-step hypothesis testing)
b. Confirm your answer using the P-value method

Answer 1

1): The null and alternative hypothesis :

Calculating test-statistic: values n 1= 40 , n2= 40 , x1= 35 ,x2= 30 , sd1= 6 ,sd2= 5

Putting the values

z= (35-30)-(0)/ 1.2349

z= 5

z-critical = 2.33 (using right tailed z critical value table )

p- value = 0.0000 (using standard normal z -distribution table )

a): Critical value decision rule : [z> 2.33] Reject the null hypothesis , bacause test-statistic falls in the rejection resion .

Conclusion : Yes, we can conclude at 0.1 alpha,that the mean working /hours week of students in LIU Beirut is higher than the mean working hours /week of students in bekaa campus .  

b): P- value decision rule : [p < alpha] Reject the null hypothesis , because p value is less than alpha, and the result is statistically significant at 0.01 alpha.