Use for #6-7. The manufacturer of a gasoline additive claims that the use of thi

Question

Use for #6-7. The manufacturer of a gasoline additive claims that the use of this additive increases gasoline mileage. A random sample of 6 cars was selected, and these cars were driven for one week without the gasoline additive and then for one week with the additive. The following table gives the mileage (in miles per gallon) for these cars with and without the additive: Car 1 2 3 4 5 6 Mileage Without Additive 24.6 28.3 18.9 23.7 15.4 29.5 Mileage With Additive 26.3 31.7 18.2 25.3 18.3 30.9

6. Assuming mileages are normally distributed, can you conclude that the use of the gasoline additive increases mileage at the 2.5% significance level?

7. Find a 90% confidence interval for the difference in mileage between cars with and without additive.

Explanation / Answer

Q#6.

Given that,
population mean(u)=18.9
sample mean, x =23.7
standard deviation, s =15.4
number (n)=29.5
null, H0: Ud > 0
alternate, H1: Ud < 0
level of significance, = 0.025
from standard normal table,left tailed t /2 =2.571
since our test is left-tailed
reject Ho, if to < -2.571
we use Test Statistic
to= d/ (S/n)
Where
Value of S^2 = [ di^2 – ( di )^2 / n ] / ( n-1 ) )
d = ( Xi-Yi)/n) = -1.717
We have d = -1.717
Pooled variance = Calculate value of Sd= S^2 = Sqrt [ 27.87-(-10.3^2/6 ] / 5 = 1.427
to = d/ (S/n) = -2.946
Critical Value
The Value of |t | with n-1 = 5 d.f is 2.571
We got |t o| = 2.946 & |t | =2.571
Make Decision
Hence Value of | to | > | t | and Here we Reject Ho
p-value :left tail - Ha : ( p < -2.9463 ) = 0.01601
hence value of p0.025 > 0.01601,here we do not reject Ho
ANSWERS
---------------
null, H0: Ud > 0
alternate, H1: Ud < 0
test statistic: -2.946
critical value: reject Ho, if to < -2.571
decision: Reject Ho
p-value: 0.01601
we have evidence that use of the gasoline additive increases mileage

Q#7.

Confidence Interval
CI = d ± t a/2 * (Sd/ Sqrt(n))
Where,
d = di/n
Sd = Sqrt( di^2 – ( di )^2 / n ] / ( n-1 ) )
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
d = ( di/n ) =-10.3/6=-1.717
Pooled Sd( Sd )= Sqrt [ 27.87- (-10.3^2/6 ] / 5 = 1.427
Confidence Interval = [ -1.717 ± t a/2 ( 0.824/ Sqrt ( 6) ) ]
= [ -1.717 - 2.015 * (0.583) , -1.717 + 2.015 * (0.583) ]
= [ -2.891 , -0.542 ]

X Y X-Y (X-Y)^2 24.6 26.3 -1.7 2.89 28.3 31.7 -3.4 11.56 18.9 18.2 0.7 0.49 23.7 25.3 -1.6 2.56 15.4 18.3 -2.9 8.41 29.5 30.9 -1.4 1.96 -10.3 27.87

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