Question

15. (8 points) In August 2012, the national average price for a gallon of regular unleaded gasoline was $3.63. The prices for a sample of n-10 gas stations in the state of Illinois were used to create the following bootstrap distribation. Bootstrap Dotplot of ean Originall Sample ร00 Bootstrap Sample 25 3.80 .5 3.90 3.95 4.00 4.05 410 41342 (a) Use the bootstrap standard error and the normal approimation to construct a 90% confidence interval for the mean price of gasoline in Illinois. (b) Interpret the coafidence interval in context (c) Does this confidence interval provide evidence that the average price in linois is different than the national average? Why?

Answer 1

a) Is it reasonable to use the t-distribution to perform a test about the average gas price in Illinois (on August 8, 2012)?

Yes, it is reasonable to use the t-distribution to perform a test about the average gas price in Illinois because we do not know the population variance.

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b) Test, at the 5% level, if there is evidence that the average gas price in Illinois (on August 8, 2012) was significiantly higher than the national average. Include all of the details of the test.

Ho: mu=3.63

Ha: mu>3.63

The test statistic is

t=(xbar-mu)/(s/vn)

=(3.975-3.63)/(0.2266/sqrt(10))

=4.81

Given a=0.05, the critical value is t(0.05, df=n-1=9)= 1.83 (from student t table)

Since t=4.81 is larger than 1.83, we reject Ho.

So we can conclude that there is evidence that the average gas price in Illinois (on August 8, 2012) was significiantly higher than the national average.

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c) Construct a 95% confidence interval for the mean gas price in Illinois ( on August 8,2012). Round your margin of error to three decimal places.

Given a=0.05, t(0.025, df=9) = 2.26 (from student t table)

So 95% confidence interval is

xbar +/- t*s/vn

--> 3.975 +/- 2.26*(0.2266/sqrt(10))

--> (3.813, 4.137)