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 47 cars in the student parking lot and find the average age to be 9.7 years with a standard deviation of 6.5 years. A random sample of 48 cars in the faculty parking lot have an average age of 3 years with a standard deviation of 3.8 years.

Note: The degrees of freedom for this problem is df = 73.845664. Round all results to 4 decimal places. Remember not to round for intermediate calculations!

1. The null hypothesis is ?0:??=??H0:μs=μf. What is the alternate hypothesis?
A. ??:??<??HA:μs<μf
B. ??:??≠??HA:μs≠μf
C. ??:??>??HA:μs>μf

2. Calculate the test statistic.  ? z t X^2 F  =

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 1.26. 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 0.029 and an effect size of 0.821. 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 contradicts the community college results because the p-value is much bigger
C. It confirms the community college results because the effect size is nearly the same.
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

1.

The aim is to see whether the cars of the students , on an average, are older than the cars of the faculty. Hence the correct alternative hypothesis is

C.

2.

	Student	Faculty
Sample Size	47	48
Mean	9.7	3
SD	6.5	3.8

Now the pooled standard deviation is calculated as

Hence the test statistic is:

3.

The df of the problem should be 47+48-2=93.

The required p-value is less than 0.00001

The given df is 73.84566474 and the p-value in that case is also less than 0.00001

Hence the null hyothesis gets rejected at 5% level of significance

4.

The correct option is A: It contradicts the community college results because the effect size is much smaller

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