Question

4) (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?

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 10281 with a standard deviation of 1813. 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. extremely strong evidence
B. some 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. ($  , $ )

The researcher decides to conduct a hypothesis test. They take a random sample of 15 such schools in Michigan and find a samp

(8 points) In 2015, the average cost of attendance at public 2-year colleges and technical schools across the United States w

Normal Q-Q Plot -2 -1 0 2 1 Theoretical Quantiles 0009 0000 000 0000 0008 Sample Quantiles

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Answer #1

Ans:

1)

Yes,as normal quantile plot is linear enough and we can assume that population is normally distributed.

2)

Option C is correct.

HO :μ = 12260

HA :μ< 12260

3)

Test statistic:

t=(10281-12260)/(1813/SQRT(15))

t=-4.2276

4)df=15-1=14

p-value=tdist(4.2276,14,1)=0.0004

5)extremely strong evidence

6)t*=tinv(0.05,14)=2.1448

margin of error=2.1448*1813/sqrt(15)=1004.0059

lower limit=10281-1004.0059=9276.9941

upper limit=10281+1004.0059=11285.0059

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