Question

A shipping firm suspects that the mean life of a certain brand of tire used by its trucks is less than 38,000 miles. To check the claim, the firm randomly selects and tests 18 of these tires and gets a mean lifetime of 37,300 miles with a standard deviation of 1000 miles. At α= 0.05, does the test suggest that mean life is less than 39000 miles?

Answer 1

Answer:

$t=\frac{37300-38000}{\frac{1000}{\sqrt{18}}}=-2.970$  

$p_v =P(t_{17}<-2.97)=0.0043$  

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v<\alpha$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is lower than 38000 at 5% of signficance.  

Step-by-step explanation:

We assume this question: A shipping firm suspects that the mean life of a certain brand of tire used by its trucks is less than 38,000 miles. To check the claim, the firm randomly selects and tests 18 of these tires and gets a mean lifetime of 37,300 miles with a standard deviation of 1000 miles. At α= 0.05, does the test suggest that mean life is less than 38000 miles?  because if is not this one is not appropiate

Data given and notation  

$\bar X=37300$ represent the sample mean    

$s=100$ represent the sample standard deviation for the sample  

$n=18$ sample size  

$\mu_o =7500$ represent the value that we want to test  

$\alpha=0.05$ represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

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

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is lower than 38000, the system of hypothesis would be:  

Null hypothesis:$\mu \geq 38000$  

Alternative hypothesis:$\mu < 38000$  

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)  

t-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".  

Calculate the statistic  

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

$t=\frac{37300-38000}{\frac{1000}{\sqrt{18}}}=-2.970$  

P-value  

The degrees of freedom are given by:

$df = n-1= 18-1=17$

Since is a left tailed test the p value would be:  

$p_v =P(t_{17}<-2.97)=0.0043$  

Conclusion  

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v<\alpha$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is lower than 38000 at 5% of signficance.