The average mpg usage for a 2009 Toyota Prius for a sample of 7 tanks of gas was
ID: 3396675 • Letter: T
Question
The average mpg usage for a 2009 Toyota Prius for a sample of 7 tanks of gas was 43.5 with a standard deviation of 1.8. For a 2009 Honda Insight, the average mpg usage for a sample of 7 tanks of gas was 39.9 with a standard deviation of 2.0.
The average mpg usage for a 2009 Toyota Prius for a sample of 7 tanks of gas was 43.5 with a standard deviation of 1.8. For a 2009 Honda Insight, the average mpg usage for a sample of 7 tanks of gas was 39.9 with a standard deviation of 2.0.
H0: toy – hon 0 vs. H1: toy – hon > 0. Reject H0 if tcalc > 1.782
H0: toy – hon 0 vs. H1: toy – hon > 0. Reject H0 if tcalc < 1.782
H0: toy – hon 0 vs. H1: toy – hon < 0. Reject H0 if tcalc < – 1.782
H0: toy – hon 0 vs. H1: toy – hon < 0. Reject H0 if tcalc > – 1.782
a-2) Calculate the tcalc. (Round your answer to 4 decimal places.)
The average mpg usage for a 2009 Toyota Prius for a sample of 7 tanks of gas was 43.5 with a standard deviation of 1.8. For a 2009 Honda Insight, the average mpg usage for a sample of 7 tanks of gas was 39.9 with a standard deviation of 2.0.
The average mpg usage for a 2009 Toyota Prius for a sample of 7 tanks of gas was 43.5 with a standard deviation of 1.8. For a 2009 Honda Insight, the average mpg usage for a sample of 7 tanks of gas was 39.9 with a standard deviation of 2.0.
Assuming equal variances, at = 0.05, is the true mean mpg lower for the Honda Insight? (a-1) Choose the appropriate hypotheses.H0: toy – hon 0 vs. H1: toy – hon > 0. Reject H0 if tcalc > 1.782
H0: toy – hon 0 vs. H1: toy – hon > 0. Reject H0 if tcalc < 1.782
H0: toy – hon 0 vs. H1: toy – hon < 0. Reject H0 if tcalc < – 1.782
H0: toy – hon 0 vs. H1: toy – hon < 0. Reject H0 if tcalc > – 1.782
Explanation / Answer
Give f(x) = (0.5) x 0<x<2
Mean = E(x) = x . f(x) dx
= x . (0.5) x dx
= 0.5 x 2 dx
= 0.5 [x3 / 3]02
= 0.5(8/3)
Mean = 1.3333 µm
in case of continuous distribution median is the point which divides the total area into two equal parts thus if M is the median then
0M f(x) dx = 1/2
0M (0.5) x dx = 1/2
0.5 0M (x) dx = 1/2
(0.5) (x2/2)0M = 1/2
(0.5) M2 = 1
M2 = 1 / 0.5
M2 = 2
Median = 1.414
Given n1 = 7 n2 = 7 mean(x1) = 43.5 mean (x2) = 39.9 s.d(s1) =1.8 s.d(s2) =2
The null hypothesis is
H0 : µtoy - µhon 0
Against the alternative hypothesis
H0 : µtoy - µhon > 0
The test statistic is
t = (( X1 - X 2) /(n1s1+n2s2)/n1+n2-2) * (1/n1)+(1/n2) ) t(n1+n2-2)
t = (( 43.5 - 39.9) /(7*1.8 + 7*2)/7+7-2) * (1/7)+(1/7) ) t(7+7-2)
t = 3.6 / 2.2167*0.2858
t= 4.5229
here ttab at 0.05 level of significance at 12 degrees of freedom
therefore tcal > ttab we reject the null hypothesis
therefore the true mean mpg is lower for the honda yes
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