(1) according to the US department of Energy\'s website www.fueleconomy.gov, my
ID: 3328041 • Letter: #
Question
(1) according to the US department of Energy's website www.fueleconomy.gov, my 1999 Nissan frontier pickup truck should get a combined EPA estimated 21 MPG. at a significance level of 0.5, is there reason to believe that my truck gets better than that?
(2) calculate and interpret a 99% confidence interval for the true average MPG for my truck.
t, test(truckSmpg , mu=21, alternative="less", conf.level .99) #CI = ## One Sample t-test ##data: truck$mpg ## t-3. 7535 , df 80 , p-value = 0 . 9998 ## alternative hypothesis: true mean is less than 21 ## 99 percent confidence interval : -Inf 23.61634 ## sample estimates: ## mean of x ## 22.60272 conf,level- .99) #ca t.test(truckSmpg, mu#21, alternatives"greater", ## One Sample t-test ##data: truck$mpg ## t = 3.7535, df = 80, p-value = 0.0001645 ## alternative hypothesis: true mean is greater than 21 ## 99 percent confidence interval: ## 21.5891 ## sample estimates: ## mean of x ## 22.60272 InfExplanation / Answer
(1)
From given codes it is. Clrearly shows that here we use on sample t test.
From above results
T statistics = 3.7525
Df=80
P value = 0.0001645 <0.5
Therefore the result is significant.
(2) 99% confidence interval is (21.47602,23.72941)
A 95% confidence interval has 0.95 probability of containing population mean.
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