A comparison of the wearing out quality of two types of tyres was obtained by road testing. Samples of 100 tyres were collected. The miles traveled until wear out were recorded and the results given were as follows

**Tyres T1 T2**

Mean x¯_{1} = 26400 miles x¯_{2} = 25000 miles

Variance S^{2}1= 1440000 miles S^{2}2= 1960000 miles

Determine a confidence interval at the confidence level of 70 percent

**Solution**

x¯_{1} = 26400

x¯_{2} = 25000

Difference between the two means

(x¯_{1} - x¯_{2}) = (26400 - 25000)

= 1,400

Again we consider the absolute value of the difference among the two means

We calculate the standard error as given below:

S_{(}_{x¯}_{A}_{ - x¯}_{B)} = √{(s^{2}_{1}/n_{1}) + (s^{2}_{2}/n_{2})}

= √{(1,440,000/100) + (1,960,000/100)}

= 184.4

Confidence level at 70 percent is read from the normal tables as 1.04 (Z = 1.04).

Thus the confidence interval is calculated as given below:

= 1400 ± (1.04) (184.4)

= 1400 ± 191.77

or (1400 - 191.77) to (1400 + 191.77)

1,208.23 ≤ X ≤ 1591.77