Question

It is advertised that the average braking distance for a small car traveling at 65 miles per hour equals 120 feet. A transportation researcher wants to determine if the statement made in the advertisement is false. She randomly test drives 36 small cars at 65 miles per hour and records the braking distance. The sample average braking distance is computed as 114 feet. Assume that the population standard deviation is 22 feet.
(1) State the null and the alternative hypotheses for the test
(2) Calculate the value of the test statistic and the p-value.

Answer 1

Answer:

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is equal to 120 or not, the system of hypothesis would be:  

Null hypothesis:$\mu = 120$  

Alternative hypothesis:$\mu \neq 120$  

Part 2

Calculate the statistic

We can replace in formula (1) the info given like this:  

$z=\frac{114-120}{\frac{22}{\sqrt{36}}}=-1.636$    

P-value

Since is a two sided test the p value would be:  

$p_v =2*P(z$  

Step-by-step explanation:

Data given and notation  

$\bar X=114$ represent the sample mean  

$\sigma=22$ represent the population standard deviation

$n=36$ sample size  

$\mu_o =120$ represent the value that we want to test

$\alpha$ represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

$p_v$ represent the p value for the test (variable of interest)  

Part 1

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is equal to 120 or not, the system of hypothesis would be:  

Null hypothesis:$\mu = 120$  

Alternative hypothesis:$\mu \neq 120$  

If we analyze the size for the sample is > 30 and we  know the population deviation so is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$  (1)  

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Part 2

Calculate the statistic

We can replace in formula (1) the info given like this:  

$z=\frac{114-120}{\frac{22}{\sqrt{36}}}=-1.636$    

P-value

Since is a two sided test the p value would be:  

$p_v =2*P(z$