Question

The Federal Clean Air Act requires that before a new fuel or fuel additive is sold in the United States, the producer must demonstrate that the emission products generated will 1 not cause a vehicle to fail to achieve compliance with certified emission standards. To estimate the difference in emission levels, the U.S. Environmental Protection Agency (EPA) requires, among other tests, a matched-pair test in which a random sample of cars is driven once with a standard fuel and then once with the new fuel. For each car, the difference emissions with new fuel emissions with standard fuel is computed. If the average difference yields a statistically significant increase in emissions with the new fuel (at the 5% significance level), the new fuel fails the test. The data below on the next page show the nitrous oxide (NOX) emissions for sixteen cars driven first once with a standard fuel and once with Petrocoal, a gasoline with a methanol additive. Note: you will most likely want to use a calculator or Excel for this problem. 

(a) (10 points) Does Petrocoal pass the test? 

(b) (10 points) Suppose you ignored the fact that the data are in the form of matched pairs, and instead treated them as two independent samples. Would Petrocoal pass the test? What is the reason for the difference in the results? Which form of the test is preferable, and why? You can assume the variance in the two populations (cars operating with Petrocoal and cars operating with standard fuel) is the same

Answer 1

Answer:-

Given That:-

The Federal Clean Air Act requires that before a new fuel or fuel additive is sold in the United States, the producer must demonstrate that the emission products generated will 1 not cause a vehicle to fail to achieve compliance with certified emission standards.

a) Does Petrocoal pass the test?

For testing if there is a significant difference between the two samples, we'll perform the paired sample t-test. Stating the hypothesis as:

Ho: Difference between the means of petrocoal and standard fuel emissions is 0

Ha: Difference between the means of petrocoal and standard fuel emissions is > 0

For performing the test, we do the below calculations:

Sample Car Petrocoal Standard Fuel Difference Difference 12 1 1.385 1.195 0.19 0.0361 2 1.23 1.185 0.045 0.002025 3 0.755 0.755 0 0 4 0.715 0.06 0.0036 0.775 2.024 5 1.805 0.219 0.047961 6 1.807 -0.015 0.000225 7 1.792 2.387 0.532 0.18 0.0324 2.207 0.301 8 0.231 0.053361 0.035344 9 0.875 0.188 10 0.541 0.687 0.498 1.843 0.043 11 2.186 0.343 12 0.809 0.838 -0.029 0.001849 0.117649 0.000841 0.0324 0.0004 0.0081 13 0.9 0.18 0.72 0.58 14 15 0.6 0.72 1.04 0.63 0.02 0.09 -0.4 1.345 16 1.44 0.16 SUM 0.532255

For calculating the T statistic we use the below formula:

(ΣD)/Ν t = ΣD2-(ΣΒ2 (N-1) (Ν)

Where, ΣD: Sum of the differences
ΣD2: Sum of the squared differences
(ΣD)2: Sum of the differences, squared

t (1.345)/16 0.532255-(1.345.61.345 (16-1)(16)

t = 2.011

Now, degree of freedom = n-1 = 16-1 = 15

With this degree of freedom and 0.05 as confidence interval, we find Tcrit which is 1.75.

Here, the calculated t-value is greater than the Tcrit.

p-value comes out to be 0.03 < 0.05. Hence, we can reject the null hypothesis.

There is a significant difference between the emissions of petrocoal and standard fuel. And, petrocoal does not pass the test.

(b) Suppose you ignored the fact that the data are in the form of matched pairs, and instead treated them as two independent samples. Would Petrocoal pass the test? What is the reason for the difference in the results? Which form of the test is preferable, and why? You can assume the variance in the two populations (cars operating with Petrocoal and cars operating with standard fuel) is the same.

If the samples are independent, then we perform the independent sample t-test. Stating the hypothesis as:

Ho: Difference between the means of petrocoal and standard fuel emissions is 0

Ha: Difference between the means of petrocoal and standard fuel emissions is > 0

For performing the test, we do the below calculations:

Step 1: Sum the two groups.
Step 2: Square the sums from Step 1
Step 3: Calculate the means for the two groups.
Step 4: Square the individual scores and then add them up.

Sample Car 1 6 7 Petrocoal Standard Fuel Petrocoal 12 Standard Fuel^2 1.39 1.20 1.92 1.43 1.23 1.19 1.51 1.401 0.76 0.76 0.57 0.57 0.78 0.72 0.60 0.51 2.02 1.81 4.101 3.26 1.79 1.81 3.21 3.27 2.39 2.21 5.70 4.87 0.53 0.30 0.28 0.09 0.88 0.69 0.77 0.47 0.54 0.50 0.29 0.25 2.19 1.84 4.78 3.401 0.81 0.84 0.65 0.70 00 9 10 11 12 13 14 0.90 0.60 0.81 0.36 0.52 0.34 15 0.52 0.72 1.04 18.55 16 0.72 0.58 0.63 1.44 17.21 296.05 1.08 0.40 2.07 1.08 27.15 23.54 SUM SUM^2 MEAN 344.14 1.16

Step 5: Insert the numbers in below formula:

ДА — ДВ t = (E42- (EA) 2 ) + (Евг (2:В)2 [ + ] пA+пв-2

where, (ΣA)2: Sum of data set A, squared (Step 2).
(ΣB)2: Sum of data set B, squared (Step 2).
μA: Mean of data set A (Step 3)
μB: Mean of data set B (Step 3)
ΣA2: Sum of the squares of data set A (Step 4)
ΣB2: Sum of the squares of data set B (Step 4)
nA: Number of items in data set A
nB: Number of items in data set B

where A is petrocoal and B is standard fuel

t = 1.16 – 1.08 (27.15– 344.4)+(23.54- 16+16-2 296.05 16 * [ib + 1]

t = 1.89

Now, degree of freedom = n-1 = 16-1 = 15

With this degree of freedom and 0.05 as confidence interval, we find Tcrit which is 1.69.

Here, the calculated t-value is greater than the Tcrit.

p-value comes out to be 0.03 < 0.05. Hence, we can reject the null hypothesis.

There is a significant difference between the emissions of petrocoal and standard fuel. And, petrocoal does not pass the test.

Both the tests give same results but we should go with paired sample t-test in this case as checking the emissions on same samples would give us more accurate picture.