Question

# Randomly selected 3131 student cars have ages with a mean of 7.37.3 years and a standard...

Randomly selected 3131 student cars have ages with a mean of 7.37.3 years and a standard deviation of 3.43.4 years, while randomly selected 2525 faculty cars have ages with a mean of 5.65.6 years and a standard deviation of 3.53.5 years.

1. Use a 0.050.05 significance level to test the claim that student cars are older than faculty cars.

(a) The null hypothesis is H0:μs=μfH0:μs=μf. What is the alternate hypothesis?

A. HA:μs>μfHA:μs>μf
B. HA:μs≠μfHA:μs≠μf
C. HA:μs<μfHA:μs<μf

(b) The test statistic is

(c) The p-value is

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

A. No
B. Yes

(a) The null hypothesis is H0:μs=μf. The alternate hypothesis is

C. HA:μs<μf

b)

n1 = 3131

n2 = 2525  s1 = 3.43

s2 = 3.53

claim : The student cars are older than faculty cars.

Null and alternative hypothesis is

H0 :u1 = u2

Vs

H1 :u1< u2

Level of significance = 0.05

Before doing this test we have to check population variances are equal or not.

Null and alternative hypothesis is Vs Test statistic is

F = Larger variance / Smaller variance = 12.4609 / 11.7649 = 1.059

Degrees of freedoms

Degrees of freedom for numerator = n1 - 1 = 3131 - 1 = 3130

Degrees of freedom for denominator = n2 - 1 = 2525 - 1 = 2524

Critical value = 1.064( using f-table )

F test statistic < critical value    we fail to reject null hypothesis.

Conclusion: Population variances are equal.

So we have to use pooled variance.

Test statistic

Formula       c) d.f = n1 + n2 – 2 = 3131 +2525 - 2 = 5654

p-value = 0.00001 ( using t table )

p-value , Reject H0

conclusion : There is a sufficient evidence to conclude that student cars are older than faculty cars

so option B is correct

B. Yes

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