Suppose μ1 and μ2 are true mean stopping distances 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 0.01 to test H0: μ1 − μ2 = −10 versus Ha: μ1 − μ2 < −10 for the following data: m = 8, x = 115.6, s1 = 5.04, n = 8, y = 129.3, and s2 = 5.32. Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to three decimal places.)

t =

P-value =

Step-by-step explanation:

This is a test of 2 independent groups. Given that μ1 and μ2 are true mean stopping distances at 50 mph for cars of a certain type equipped with two different types of braking systems, the hypothesis are

For null,

H0: μ1 − μ2 = - 10

For alternative,

Ha: μ1 − μ2 < - 10

This is a left tailed test.

Since sample standard deviation is known, we would determine the test statistic by using the t test. The formula is

(x1 - x2)/√(s1²/n1 + s2²/n2)

From the information given,

x1 = 115.6

x2 = 129.3

s1 = 5.04

s2 = 5.32

n1 = 8

n2 = 8

t = (115.6 - 129.3)/√(5.04²/8 + 5.32²/8)

t = - 2.041

Test statistic = - 2.04

The formula for determining the degree of freedom is

df = [s1²/n1 + s2²/n2]²/(1/n1 - 1)(s1²/n1)² + (1/n2 - 1)(s2²/n2)²

df = [5.04²/8 + 5.32²/8]²/[(1/8 - 1)(5.04²/8)² + (1/8 - 1)(5.32²/8)²] = 45.064369/3.22827484

df = 14

We would determine the probability value from the t test calculator. It becomes

p value = 0.030

Since alpha, 0.01 < the p value, 0.03, then we would fail to reject the null hypothesis.

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step-by-step explanation:

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