Question

Randomly selected 23 student cars have ages with a mean of 8 years and a standard deviation of 3.6 years, while randomly selected 27 faculty cars have ages with a mean of 5.8years and a standard deviation of 3.7 years.

1. Use a 0.01significance level to test the claim that student cars are older than faculty cars.

(a) The test statistic is

(b) The critical value is

(c) Is there sufficient evidence to support the claim that student cars are older than faculty cars?

A. Yes
B. No


2. Construct a 99% confidence interval estimate of the difference ?s??f, where ?s is the mean age of student cars and ?f is the mean age of faculty cars.

< (?s??f) <

Answer 1

a)

Let x1 is the mean age of student cars, x2 is the mean age of faculty cars,  s1 is the standard deviation of age of student cars, x2 is the standard deviation of age of faculty cars and SE is the standard error.

Given,

x1 = 8 , s1 = 3.6, n1 = 23

x2 = 5.8, s2 = 3.7, n2 = 27

SE = sqrt[ (s1²/n1) + (s2²/n2) ]

= sqrt[ (3.6²/23) + (3.7²/27) ]

= 1.034657

Test statistic, t = (x1 - x2) / SE = (8 - 5.8) / 1.034657 = 2.126309

(b)

Degree of freedom is given as,

DF = (s1²/n1 + s2²/n2)² / { [ (s1² / n1)² / (n1 - 1) ] + [ (s2² / n2)² / (n2 - 1) ] }

= 47 (Rounding to nearest integer)

Critical value of t at 0.01significance level (one tail test) and degree of freedom as 47 is 2.41

(c)

As, the test statistic (2.126309) is less than the critical value, we fail to reject the null hypothesis and conclude that there is not sufficient evidence to support the claim that student cars are older than faculty cars. The answer is B. No

2.

Critical value of t at 0.01significance level (one tail test) and degree of freedom as 47 is 2.41

99% confidence interval estimate of the difference ?s??f, where ?s is the mean age of student cars and ?f is the mean age of faculty cars is,

(x1 - x2) - t * SE < (?s??f) < (x1 - x2) + t * SE

(8 - 5.8) - 2.41 * 1.034657 < (?s??f) < (8 - 5.8) + 2.41 * 1.034657

-0.2935 < (?s??f) < 4.6935