N-1

A
A2
B2

B
difference ,
diff
two sided alt. Hyp.
one sided alt. Hyp.
1.
(7pts)(Hypothetical) A company claims that a gas additive it produces increases gas mileage. To test this claim, 36 (36) identical model cars were filled with 10 gallons of gas. These cars were then run on a test course under identical driving conditions until they had only one half of a gallon of gas remaining (indicated by a sensor in the gas tank), at which time they recorded the mileage. The cars then had their gas tanks drained and refilled with 10 gallons of gas which contained one can of the gas additive. Again, the cars ran on the test course as before and recorded their mileage when only one half of a gallon remained in the tank. In the first run, with gas alone, the cars had an average mileage of 224 with an SD= 13. In the second run, the average mileage was 230 with an SD= 10. Is there enough evidence to support the company’s claim?
Test the hypothesis H0:

Additive
<=

NoAdditive
H1:

Additive
>

NoAdditive
(you must include a test statistic, p value and conclusion).
Ans: Since n=36 for both samples, we can use a Z-test. While the issue here is about gas mileage,you don’t need to calculate mpg here, since the mileage measurements are from identical conditions, and hence are an indirect measure of mpg (a higher mileage indicates a higher mpg). The test statistic (using the formula above) is
Z
TEST
= [(230-224)-0]/Sqrt{ [10/Sqrt(36)] 2 + [13/Sqrt(36)] 2 }
= 6/Sqrt( 1.67
2 + 2.17
2 )
=2.19 from which we get a p-value of 1-.986= .014<.05. So we reject H0 and conclude that the average mileage of the tested cars increases with the addition of the fuel additive.
2.
(4pts)A comparison of two groups , A and B, was conducted using two samples. Sample one, from A, had a sample size of 20. Sample two, from B, had a sample size of 26. Given H0:

A =

B , H1:

A≠

B, a significance level of 0.05, and a test statistic for the difference t= 2.0, find a p value and state whether you accept or reject H0
Ans: The first question here is what are your degrees of freedom? Since at least one sample is less than 30, you know you will use the t-test. Here, df = the smaller of nA-1 and nB-1, in this case it will be nA-1=19. So for t=2.0 with df=19, we get a .05>(right tailed)p value >.025. However, since the original hypothesis is two sided, i.e.

A =

B vs. H1:

A≠

B, the actual p value we use to make our conclusion is 2*(right tailed p value), so we arrive at
.1> two sided p value >.05. Consequently, we don’t have enough evidence to reject, and hence we accept H0.

3.
A education specialist hypothesizes that scores on a particular skills tests are approx. normally distributed with a mean of 65 and a standard deviation of 10. In a random sample of 100 , the scores are distributed as follows:
Tests score: Below 55 55-65 65-75 above 75
# of people: 11 35 40 14
Using a
 2 test , test the hypothesis that the distribution is normally distributed (Recall that 68% of the data is supposed to be (symmetrically) within 1 SD of the mean, and
32% of the data is supposed to be (symmetrically) outside the interval mean +- 1SD).
Your should have four terms in your sum.
Ans: To be formal,
H0: test scores are normally distributed (that is, 68% of the data is supposed to be (symmetrically) within 1 SD of the mean, and 32% of the data is supposed to be
(symmetrically) outside the interval mean +- 1SD).
H1: test scores are not normally distributed
The range 55-65 is equivalently from mean-1SD to mean. Similarly, the range 65 -75 is the range mean to mean+ 1SD. So, in the overall range of 55-75, i.e. mean +- 1SD, we expect to see 68% of the sample, or 68 test scores, if the specialist’s hypothesis is correct.
Since this should be divided up symmetrically, we expect to see 34 people in each of the ranges 55-65 and 65-75. A similar comparison indicates we expect, if the specialist’s hypothesis is correct, we expect to see 16 people in each of the ranges below 55 and above 75. So our test statistic is
 2 test = (11-16) 2 /16 + (35-34) 2 /34 + (40-34) 2 /34 + (14-16) 2 /16 = 2.9
Since there are four categorical “responses”, our df=4-1 = 3. With

2 test=2.9 and df=3, we get a p value>.2 >.05, so we cannot reject H0,i.e. the hypothesis that the test scores
are normally distributed.