Question

Cars on Campus. Statistics students at a community college wonder whether the cars belonging to students are, on average, older than the cars belonging to faculty. They select a random sample of 18 cars in the student parking lot and find the average age to be 8.4 years with a standard deviation of 7.3 years. A random sample of 15 cars in the faculty parking lot have an average age of 4.2 years with a standard deviation of 3.8 years.

1. The null hypothesis is H0:????????=???????? H 0 : μ s = μ f . What is the alternate hypothesis? A. H????:????????≠???????? H A : μ s ≠ μ f B. H????:????????<???????? H A : μ s < μ f C. H????:????????>???????? H A : μ s > μ f

2. Calculate the test statistic. =

3. Calculate the p-value for this hypothesis test. p value =

4. Suppose that students at a nearby university decide to replicate this test. Using the information from the community college, they calculate an effect size of 0.7. Next, they obtain samples from the university student and faculty lots and, using their new sample data, conduct the same hypothesis test. They calculate a p-value of -0.0213 and an effect size of 0.487. Do their results confirm or conflict with the results at the community college? A. It contradicts the community college results because the effect size is much smaller. B. It confirms the community college results because the effect size is nearly the same. C. It contradicts the community college results because the p-value is much bigger D. It can neither confirm or contradict the community college results because we don't know the sample sizes the university students used. E. It confirms the community college results because the p-value is much smaller.

Answer 1

Let the 1st group be of students and 2nd group be of faculty.

n1 = 18

\bar{x}_{1} = 8.4

s1 = 7.3

n2 = 15

\bar{x}_{2} = 4.2

s2 = 3.8

1.

H0: =

HA: >

2.

Equality of variance test:

F = s12/s22

= 53.29/14.44

= 3.69

df-num = n1-1 = 17

df-den = n2-1 = 14

p-value = 0..008 < 0.05 i.e.H0 can be rejected and hence we can say that both populations do not have the same variances.

So, test-statistic will be calculated as follows:

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3.

df = 26

p-value = 0.0217

4.

E. It confirms the community college results because the p-value is much smaller.