Question

The average work week for engineers in a start-up company is believed to be about 60 hours. A newly hired engineer hopes that it's shorter. She asks 10 engineering friends in start-ups for the lengths of their average work weeks. Based on the results that follow, should she count on the average work week to be shorter than 60 hours? Conduct a hypothesis test at the 5% level.

Data (length of average work week): 70; 45; 55; 60; 65; 60; 60; 60; 55; 55.

State the distribution to use for the test. (Enter your answer in the form z or t_df where df is the degrees of freedom.) ????

Answer 1

	Values ( X )	Σ ( Xi- X̅ )2
	70	132.25
	45	182.25
	55	12.25
	60	2.25
	65	42.25
	60	2.25
	60	2.25
	60	2.25
	55	12.25
	55	12.25
Total	585	402.5

Mean X̅ = Σ Xi / n
X̅ = 585 / 10 = 58.5
Sample Standard deviation S_X = √ ( (Xi - X̅ )2 / n - 1 )
S_X = √ ( 402.5 / 10 -1 ) = 6.6875

Since the population standard deviation is unknown we use t test.

To Test :-

H0 :- µ = 60
H1 :- µ < 60

Test Statistic :-
t = ( X̅ - µ ) / (S / √(n) )
t = ( 58.5 - 60 ) / ( 6.6875 / √(10) )
t = -0.7093

Test Criteria :-
Reject null hypothesis if t < -t_{(α, n-1)}

DF = n - 1 = 10 - 1 = 9

t_{(α, n-1)} = t_{(0.05 , 10-1)} = 1.833
t > -t_{(α, n-1)} = -0.7093 > - 1.833
Result :- Fail to reject null hypothesis

Decision based on P value
P - value = P ( t > 0.7093 ) = 0.2481
Reject null hypothesis if P value < α = 0.05 level of significance
P - value = 0.2481 > 0.05 ,hence we fail to reject null hypothesis
Conclusion :- Fail to reject null hypothesis

There is insufficient evidence to support the claim that average work week to be shorter than 60 hours.