Question

Suppose ui and M2 are true mean stopping distances (in feet) at 50 mph for cars of a certain type equipped with two different types of braking systems. Use the two-sample t test at significance level .01 to test: Ho: M1 - U2 = -10 Car Type 1: m = 6, X = 115.7, S1 = 5.03 Ha:: M1-M2 < -10 Car Type 2: n = 6, y = 129.3, and s2 = 5.38 Provide an explanation for the result of the test in terms of Car 1 and Car 2's stopping distances.

Answer 1

The provided sample means are shown below:

\$bar X_1 = 115.7 ,\bar X_2 = 129.3$

Also, the provided sample standard deviations are:

and the sample sizes are n_1 = 6 and n_2 = 6

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ1​ - μ2​ = -10

Ha: μ1​ - μ2​ < -10

This corresponds to a left-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.

(2) Rejection Region

Based on the information provided, the significance level is α=0.01, and the degrees of freedom are df = 10. In fact, the degrees of freedom are computed as follows, assuming that the population variances are equal:

Hence, it is found that the critical value for this left-tailed test is t_c = -2.764 , for α=0.01 and df = 10

(3) Test Statistics

Since it is assumed that the population variances are equal, the t-statistic is computed as follows:

$t = \frac{\bar X_1 - \bar X_2-\mu}{\sqrt{ \frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1+n_2-2}(\frac{1}{n_1}+\frac{1}{n_2}) } }$

$t = \frac{ 115.7 - 129.3-(-10))}{\sqrt{ \frac{(6-1)5.03^2 + (6-1)5.38^2}{ 6+6-2}(\frac{1}{ 6}+\frac{1}{ 6}) } }$

= -1.735

(4) Decision about the null hypothesis

Since it is observed that t = -1.735 > t_c = -2.764, it is then concluded that the null hypothesis is not rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is rejected. Therefore, there is not enough evidence to claim that population mean μ1​- μ2​ is less than -10, at the 0.01 significance level.

The difference in stopping distance of two cars is - 10