Statistics question

Question

An investigation has been carried out to compare the carbon monoxide emission levels (grams per mile) of two types of passenger cars. Twenty-five Model-A passenger cars and thirteen Model-B passenger cars were selected at random and the amount of carbon monoxide emitted from each car (cruising at 50 mph) was measured. Assume normality of the data.

For Model A: n = 25, Σx = 495, Σx^2 = 9921

For Model B: n= 13, Σx = 208, Σx^2 = 3472

(a) Test the hypothesis, at the 0.05 level of significance, that the standard deviation of the amount of carbon monoxide emitted from Model-A passenger cars is less than 2.4 grams per mile.

(b) At the 0.10 level of significance, test whether the variances of the carbon monoxide emission levels for the two models of cars are equal or not.

x is the amount of carbon monoxide emitted.

THANK YOU!!

Answer 1

Model A: mean, m=∑X/n=495/25=19.8, where n=25; and variance v is calculated as follows. 

nv=∑(X-m)²=∑X²-2(∑X)m+nm²=∑X²-2nm²+nm²=∑X²-nm².

Therefore v=∑X²/n-m²=9921/25-19.8²=4.8. Standard deviation s=√v=√4.8=2.19 approx. grams per mile.

Model B: mean=208/13=16, v=3472/13-16²=11.077, s=3.33 approx. grams per mile.

a) Degrees of freedom for Model A = 24 (one less than the size of the sample, 25). We can apply the chi squared test using (24)(2.19/2.4)²=20. The critical value from tables is 36.42. So the value of 20 is less than this, and the null hypothesis is not rejected, that is, for Model A, the sample is indeed less than 2.4 g/mile within a significance level of 0.05.

b) We may need to recalibrate the variances of the two models by using n-1 instead of n. For Model A, nv=120 and for Model B it nv=144, so the revised variances are 120/24=5 and 144/12=12. We have a 0.10 significance level this time. We also need to decide which statistical test is most suitable in this case.

On the assumption of normality, we can estimate the population standard deviation:

=√((120+144)/(24+12))=√(264/36)=√(22/3)=2.708.

We also need to know the standard error, so we multiply this by √(1/25+1/13)=0.3419, and the standard error is 0.926. Now we work out a Z-score (T value) for the difference in means: (19.8-16)/0.926=3.8/0.926=4.1 approx.

The degrees of freedom are combined: 24+12=36.

The critical t value for 36 dof at 0.10 significance level is 1.306.

If we use the difference in variance, we get (3.33-2.19)/0.926=1.23 which is lower than the critical value, so we would deduce that the null hypothesis is not rejected: the variances are the same within the 0.10 level of significance.