Question

A metropolitan transportation authority has set a bus mechanical reliability goal of 3,900 bus miles. Bus mechanical reliability is measured specifically as the number of bus miles between mechanical road calls. Suppose a sample of 100 buses resulted in a sample mean of 3,950 bus miles and a sample standard deviation of 225 bus miles.
Complete parts​ (a) and​ (b) below.
(a) Is there evidence that the population mean bus miles is more than 3,700 bus​ miles? (Use a 0.10 level of​significance.)
(b) Find the test statistic for this hypothesis test. (Round to two decimal places as​needed.) What is the p-value?

Answer 1

Answer:

a) We need to conduct a hypothesis in order to check if the true mean is more than 3900 bus miles, the system of hypothesis would be:  

Null hypothesis:$\mu \leq 3900$  

Alternative hypothesis:$\mu > 3900$  

If we compare the p value and the significance level given $\alpha=0.1$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 3900 bis miles at 10% of signficance.  

b) $t=\frac{3950-3900}{\frac{225}{\sqrt{100}}}=2.22$    

$p_v =P(t_{(99)}>2.22)=0.014$  

Step-by-step explanation:

Data given and notation  

$\bar X=3950$ represent the sample mean

$s=225$ represent the sample standard deviation

$n=100$ sample size  

$\mu_o =3900$ represent the value that we want to test

$\alpha=0.1$ 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)  

Assuming this question for part a: Is there evidence that the population mean bus miles is more than 3,700 bus​ miles? (Use a 0.10 level of​significance.) Because is if 3700 not makes sense based on the previous info

Part a: State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is more than 3900 bus miles, the system of hypothesis would be:  

Null hypothesis:$\mu \leq 3900$  

Alternative hypothesis:$\mu > 3900$  

If we analyze the size for the sample is > 30 but we don't know the population deviation so 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".  

Part b: Calculate the statistic

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

$t=\frac{3950-3900}{\frac{225}{\sqrt{100}}}=2.22$    

P-value

The first step is calculate the degrees of freedom, on this case:  

$df=n-1=100-1=9$  

Since is a one side test the p value would be:  

$p_v =P(t_{(99)}>2.22)=0.014$  

Conclusion  

If we compare the p value and the significance level given $\alpha=0.1$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 3900 bis miles at 10% of signficance.