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 a = 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

A. confidence interval

$\bar{X}_{1}-\bar{X}_{2}\pm 1.645\sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}}$

[-1.06472,2.9047]

90% confidence interval for difference between the average age of student's car and average age of faculty car is between -1.06472 and 2.9047.

B.$H_{0}$ :There is no enough average to conclude that an average faculty cars are newer than students cars.

$H_{1}$ :There is enough average to conclude that an average faculty cars are newer than students cars.

alpha=0.05,$\frac{\bar{X}_{1}-\bar{X}_{2}}{\sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}}}$,

t statistics=

pvalue=0.224955

Since pvalue is greater than the alpha, the null hypothesis may be accepted. There is no enough average to conclude that an average faculty cars are newer than students cars.