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.   

Normal Q-Q Plot O 2 -1 0 1 2 Theoretical Quantiles 0009 0000 0000 0000 0008 Sample Quantiles

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 10602 with a standard deviation of 1827. 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. H0H0: μμ = 12,260 vs. HAHA: μμ < 12,260
B. H0H0: μμ = 12,260 vs. HAHA: μ≠μ≠ 12,260
C. H0H0: μμ = 12,260 vs. HAHA: μμ > 12,260

3. Calculate the test statistic.

4. Calculate the p-value.

5. Based on the p-value, we have:
A. some evidence
B. extremely strong evidence
C. very strong evidence
D. little evidence
E. 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. ($  , $ )

Answer 1

The statistical software output for this problem is:

One sample T summary hypothesis test: u: Mean of population Ho : = 12260 HA: < 12260 Hypothesis test results: Mean Sample Mean Std. Err. DF T-Stat P-value 10602 471.72937 14 -3.5147271 0.0017

95% confidence interval results: Mean Sample Mean Std. Err. DF L. Limit U. Limit 10602 471.72937 14 9590.2411 11613.759

Hence,

1. Yes; The population data distribution seems approximately normal.

2. Option A is correct.

3. Test statistic = -3.5147

4. p - Value = 0.0017

5. very strong evidence

6. 95% confidence interval:

($ 9590.2411, $ 11613.7590)