1 State the null and alternative hypotheses 2 Choose the sta

1. State the null and alternative hypotheses.

2. Choose the statistical test.

3. Select the desired level of significance.

4. Calculate the value of the test statistic. • Show all your work for calculations that must be done \"by hand\". • show the relevant output tables.

5. Determine the critical value for the test statistic.

6. Interpret the results.

7. Calculate the effect size, if applicable

8. calculate the power.

8. Write your interpretation in APA format.

A major oil company has developed a new gasoline additive that is supposed to increase mileage. To test that hypothesis, 10 cars are randomly selected. Each car sampled is driven both with and without the additive. The resulting gas mileages, in miles per gallon (mpg), are displayed in the table below. Does the additive significantly increase mileage?

Car

With additive

Without additive

1

25.7

24.9

2

20.0

18.8

3

28.4

26.7

4

13.7

12.0

5

18.8

17.8

6

12.5

13.3

7

28.4

27.8

8

10.1

8.2

9

23.1

22.6

10

11.9

10.4

Solution

Set Up Hypothesis
Null Hypothesis , There Is No-Significance between them Ho: u1 < u2
Alternate Hypothesis, There Is Significance between them - H1: u1 > u2
Test Statistic
X(Mean)=19.26
Standard Deviation(s.d1)=6.982 ; Number(n1)=10
Y(Mean)=18.25
Standard Deviation(s.d2)=7.09; Number(n2)=10
we use Test Statistic (t) = (X-Y)/Sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =19.26-18.25/Sqrt((48.74832/10)+(50.2681/10))
to =0.32
| to | =0.32
Critical Value
The Value of |t | with Min (n1-1, n2-1) i.e 9 d.f is 1.833
We got |to| = 0.32097 & | t | = 1.833
Make Decision
Hence Value of |to | < | t | and Here we Do not Reject Ho
P-Value:Right Tail -Ha : ( P > 0.321 ) = 0.37778
Hence Value of P0.05 < 0.37778,Here We Do not Reject Ho

With additive significantly increase mileage