Question

a company is trying to decide which of two types of tires to buy for their trucks. they would like to adopt brand C unless there is some evidence that Brand D is better. An experiment was conducted where 16 tires from each brand where used. The tires were run under similar conditions until they wrote out. The result are: Brand C: Mean = 40000 kms standard deviation : 5400kms
Brand D : Mean= 38009kms, standard deviation = 3200kms

Answer 1

$n_1=n_2=16 \\ \bar{x_1}=40000 \\ s_1 = 5400 \\ \bar{x_2}= 38009 \\ s_2=3200 \\ H_0: \mu_1 \leq \mu_2 \\ H_1: \mu_ 1> \mu_2$

When n1 and n2 are small and $\sigma_1$ and $\sigma_2$ are unknown, we have to use the two-sample t test for independent random samples from two normal populations having the same unknown variance.

Test-statistic:

$t= \frac{\bar{x_1} - \bar{x_2}}{s_p \sqrt{(1/n_1) + (1/n_2)}} \\ s^2_p= \frac{(n_1-1)s^2_1 +(n_2-1)s^2_2}{n_1+n_2-2} \\ s^2_p= \frac{(16-1)(5400)^2 +(16-1)(3200)^2}{16+16-2} \\ = \frac{4.374 \times 10^8 + 1.536 \times 10^8}{30} \\ = 0.197 \times 10^8 \\ s_p=0.4438 \times 10^4 \\ = 4438.4 \\ t= \frac{40000 -38009}{4438.4 \sqrt{(1/16) + (1/16)}} \\ = \frac{1991}{1569.2} \\ = 1.268$

Reject H₀ if $t≥t_{crit}$

Let use α=0.05

For one-tail test and degree of freedom $df = n_1+n_2-2= 30, \; t_{crit} = 1.697,$

$t=1.268 < t_{crit}= 1.697.$

Accept H₀ at 0.05 significance level.

Therefore, there is enough evidence to claim that the population mean μ₁ is less than the population mean μ₂. Brand D is better than C.