Question

A sample of 10 diesel trucks were run both hot and cold to estimate the difference in fuel economy. The results, in mi/gal, are as follows:

Truck	1	2	3	4	5	6	7	8	9	10
Hot	4.56	4.46	6.49	5.37	6.25	5.90	4.12	3.85	4.15	4.69
Cold	4.26	4.08	5.83	4.96	5.87	5.32	3.92	3.69	3.74	4.19

A) Find a 98% two-sided confidence interval for the difference in mean fuel mileage between high and cold engines.

B) What probability distribution is the confidence interval based upon? How many degrees of freedom are involved?

Answer 1

A)

Here we have paired observations so paired t test will be used.

Let d = hot -cold

Following table shows the calculations:

Hot	Cold	d=hot-cold	(d-mean)^2
4.56	4.26	0.3	0.009604
4.46	4.08	0.38	0.000324
6.49	5.83	0.66	0.068644
5.37	4.96	0.41	0.000144
6.25	5.87	0.38	0.000324
5.9	5.32	0.58	0.033124
4.12	3.92	0.2	0.039204
3.85	3.69	0.16	0.056644
4.15	3.74	0.41	0.000144
4.69	4.19	0.5	0.010404
Total		3.98	0.21856

Sample size: n =10

So,

$\bar{d}=\frac{\sum d}{n}=\frac{3.98}{10}=0.398$

$s_{d}=\sqrt{\frac{\sum \left ( d-\bar{d} \right )^{2}}{n-1}}=\sqrt{\frac{0.21856}{9}}=0.1558$

The degree of freedom: df=n-1=9

The critical value of t for 98% confidence interval is: 2.8214

So required confidence interval will be

$\bar{d}\pm t_{critical}\cdot \frac{s_{d}}{\sqrt{n}}=0.398\pm 2.8214\cdot \frac{0.1558}{\sqrt{10}}=0.398\pm 0.139$

So required confidence interval is (0.259, 0.537).

(b)

t-distribution

Degree of freedom: df=n-1=9