Complete Question
The government believes that higher-octane fuels result in better gas mileage for their car. To test this claim, a researcher randomly selected 11 individuals to participate in the study. Each participant received 10 gallons of gas (87-octane and 92-octane) and drove his car on a closed course. The number of miles driven until the car ran out of gas was recorded. A coin flip was used to determine whether the car was filled up with 87-octane first or92-octane first, and the driver did not know which fuel was in the tank.
We conducted a paired t-test to see if higher-octane fuels result in better gas mileage, and the test statistic was found to 1.74.
Calculate the p-value.
Answer:
The p-value is [tex]p-value = 0.0562[/tex]
Step-by-step explanation:
From the question we are told that
The sample size is n = 11
The test statistics is [tex]t = 1.74[/tex]
The null hypothesis is [tex]H_o : \mu _1 = \mu_2[/tex]
The alternative hypothesis is [tex]H_a : \mu _1 > \mu_2[/tex]
Here [tex]\mu_1[/tex] is the gas millage for 87-octane fuel while
Here [tex]\mu_2[/tex] is the gas millage for 92-octane fuel
Generally the degree of freedom is mathematically represented as
[tex]df = n- 1[/tex]
[tex]df =10[/tex]
Generally from the student t -distribution table the probability of t = 1.74 at a degree of freedom of [tex]df =10[/tex] is
[tex]p-value = P( t > 1.74) = 0.0562[/tex]
