Question

(8 points) 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? ? becau: ✓ ? 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. Noms - Plot MONGS 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 10393 with a standard deviation of 1590. 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. HON = 12,260 vs. HA 12,260 B. HO: 4 = 12,260 vs. HAH > 12,260 C. HON = 12,260 VS. HAN < 12,260 3. Calculate the test statistic. 4. Calculate the p-value. 5. Based on the p-value, we have: A. strong evidence B. little evidence C. extremely strong evidence D. some evidence E. very 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. (S

Answer 1

(1) Q-Q plot forms a linear line which means that the data follows approximately normal distribution.

Yes, (we can conduct the test) because the data is approximately normal.

Even thought we require large size >30 but that is the case if we have the unknown distribution of population and we assume to be normal. But here the plot is showing the distribution.

$n=15\;\bar{x}=10393\;S_{x}=1590$

(2) We are testing whether the average cost is less than the national avg. A left 1-tailed test.

$\mathbf{H_{0}:\mu=12260\;vs\;H_{1}=\mu<12260}$

(3) Test Statistic: $\frac{\bar{x}-\mu_{0}}{S_{x}/\sqrt{n}}$

= $\frac{10393-12260}{1590/\sqrt{15}}$

Test Statistic = $\mathbf{-4.5477}$

We are going to use t-dist since we have unknown population standard deviation and it is 1-tailed test.

(4) p - value = P ($t_{n-1}$ > | Test Stat| )

= P ($t_{14}$ > 4.5477)

= $\mathbf{0.00023}$ (using t-dist tables)

(5) We are using a 0.05 level of significance

Since p-value < 0.05

We have

Strong evidence to reject the null hypothesis

We have sufficient or strong evidence terms in hypothesis testing. Strong because p-value < 0.01

(6) (1- $\alpha$) confidence interval for population mean is

$(\bar{x}\bar{&plus;}t_{n-1,\alpha/2}\frac{S_{x}}{\sqrt{n}})$

$t_{n-1,\alpha/2}=t_{14,0.05/2}$ ($\alpha$ = 1 - 0.95 = 0.05)

$t_{14,0.025}=2.1448$ (using t-dist tables)

Substituting the values

$(10393\bar{&plus;}2.1448\frac{1590}{\sqrt{15}})$

$\mathbf{(9512.487,11273.51)}$