Answer:
We conclude that the car have mileage rating less than 30 miles per gallon.
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 30 miles/gallon
Sample:
26, 24, 20, 25, 27, 25, 28, 30, 26, 33
[tex]\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}[/tex]
where [tex]x_i[/tex] are data points, [tex]\bar{x}[/tex] is the mean and n is the number of observations.
[tex]Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}[/tex]
[tex]Mean =\displaystyle\frac{264}{10} = 26.4[/tex]
Sum of squares of differences = 110.4
[tex]S.D = \sqrt{\frac{110.4}{9}} = 3.50[/tex]
Sample size, n = 10
Alpha, α = 0.05
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu \geq 30\text{ miles per gallon}\\H_A: \mu < 30\text{ miles per gallon}[/tex]
We use one-tailed t test to perform this hypothesis.
Formula:
[tex]t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }[/tex]
Putting all the values, we have
[tex]t_{stat} = \displaystyle\frac{26.4 - 30}{\frac{3.50}{\sqrt{10}} } = -3.25[/tex]
Now,
[tex]t_{critical} \text{ at 0.05 level of significance, 9 degree of freedom } =-1.833[/tex]
Since,
[tex]t_{stat} < t_{critical}[/tex]
We fail to accept the null hypothesis and reject it.
We conclude that the car have mileage rating less than 30 miles per gallon.
b) We assumed that the level of significance is 0.05 that is 5%.
Answer:
The car can be driven 30 miles for every gallon of gasoline used.
So, the graph that represents the fuel efficiency of the car has a slope of 30.
Therefore, graph W represents the fuel efficiency of the car.
Step-by-step explanation:
this can help someone who had this question: A car is advertised as having a fuel efficiency of 30 miles per gallon when driving on a highway. Which graph represents the fuel efficiency of this car?
