Question

An advocacy group claims that the mean braking distance of a certain type of tire is 75 feet when the car is going 40 miles per hour. In a test of 80 of these tires, the braking distance has a mean of 77 and a standard deviation of 5.9 feet. Find the standardized test statistic and the corresponding p-value.

Answer 1

Answer:

The calculated value t = 3.032 >  1.9904 at 5% level of significance

Step-by-step explanation:

Step(i):-

Given mean of the population 'μ'=75 feet

Given mean of the sample(x⁻) = 77

Given standard deviation of the sample(S) = 5.9

Given sample size 'n' =80

Step(ii):-

Test statistic

              $t=\frac{x^{-} -mean}{\frac{S}{\sqrt{n} } }$

             $t=\frac{77-75}{\frac{5.9}{\sqrt{80} } }$

            t = 3.032

Degrees of freedom ν=n-1 = 80-1 =79

tabulated value = 1.9904 at 5% level of significance

The calculated value t = 3.032

Final answer:-

The calculated value t = 3.032 >  1.9904 at 5% level of significance