Question

In 2015, the average cost of attendance at public 2-year colleges and technical schools across the United States was $12,260. A researcher wants to know if the 2-year colleges in Michigan are less expensive.

1. A QQ-plot of the cost of attendance at public 2-year colleges and technical schools is shown below. Based on this plot, is it reasonable to use this data to evaluate the research question? Why or why not?

? Yes No  because  ? The population data distribution seems approximately normal. The population data distribution does not seem approximately normal. A sample size of 15 is too small. A sample size of 15 is very large.  

The researcher decides to conduct a hypothesis test. They take a random sample of 15 such schools in Michigan and find a sample average of 10173 with a standard deviation of 1536. Conduct a hypothesis test to determine if the average cost of attendance at 2-year colleges in Michigan is lower than the national average. Round all numeric results to 4 decimal places.

2. Which set of hypotheses should the researcher use?
A. ?0H0: ?μ = 12,260 vs. ??HA: ?≠μ≠ 12,260
B. ?0H0: ?μ = 12,260 vs. ??HA: ?μ < 12,260
C. ?0H0: ?μ = 12,260 vs. ??HA: ?μ > 12,260

3. Calculate the test statistic.

4. Calculate the p-value.

5. Based on the p-value, we have:
A. little evidence
B. some evidence
C. strong evidence
D. very strong evidence
E. extremely strong evidence
that the null model is not a good fit for our observed data.

6. Calculate a 95% confidence interval for the mean cost of attendance at 2-year colleges in Michigan. ($  , $ )

Help Entering Answers

Answer 1

1)
Yes because The population data distribution seems approximately normal.
2)

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: μ = 12260
Alternative Hypothesis, Ha: μ < 12260

3)

Test statistic,
t = (xbar - mu)/(s/sqrt(n))
t = (10173 - 12260)/(1536/sqrt(15))
t = -5.2623

4)

P-value Approach
P-value = 0.0001

5)

E. extremely strong evidence

6)
sample mean, xbar = 10173
sample standard deviation, s = 1536
sample size, n = 15
degrees of freedom, df = n - 1 = 14

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, tc = t(α/2, df) = 2.145

ME = tc * s/sqrt(n)
ME = 2.145 * 1536/sqrt(15)
ME = 850.693

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (10173 - 2.145 * 1536/sqrt(15) , 10173 + 2.145 * 1536/sqrt(15))
CI = (9322.3070 , 11023.6930)