Question

7. A group of university students are interested in comparing the average age of cars owned by students and the average age of cars owned by faculty. They randomly selected 25 cars that are own by students and 20 cars that are owned by faculty. The average age and standard deviation obtained from the students’ cars are 6.78 years and 5.21 years, respectively. The sample of faculty cars produced a mean and a standard deviation of 5.86 years, and 2.72. A. Construct and interpret a 90% confidence interval for the difference between the average age of students’ cars and average age of faculty cars. B. At α = 0.05, is there enough average to conclude that on average faculty cars are newer than students’ cars? Use the p-value method.

Answer 1

Part a)

Confidence interval :-

DF = 37

t(α/2, DF) = t(0.1 /2, 37 ) = 1.687

Lower Limit =
Lower Limit = -1.1154
Upper Limit =
Upper Limit = 2.9554
90% Confidence interval is ( -1.1154 , 2.9554 )

Part b)

To Test :-

µ1 :- Average age of students cars

µ2 :-   Average age of faculty cars

H0 :- µ1 <= µ2
H1 :- µ1 > µ2

Test Statistic :-


t = 0.7625

Decision based on P value
P - value = P ( t > 0.7625 ) = 0.2253
Reject null hypothesis if P value < α level of significance
P - value = 0.2253 > 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 faculty cars are newer than students’ cars.