Question

A commonly cited standard for one-way length (duration) of school bus rides for elementary school children is 30 minutes. A local government office in a rural area randomly samples 100 elementary school children in their district and find an average one-way commute time of 38 minutes with a standard deviation of 8 minutes. Which of the following is the correct set of hypotheses for testing if the average commute time of elementary school students in this district is different than the commonly cited standard of 30 minutes? H0: x = 38; HA: μ = 30

H0: μ = 30; HA: μ ≠ 30

H0: μ = 30; HA: x = 38

HA: μ = 30; H0: x ≠ 30

H0: μ = 30; HA: μ > 30

Answer 1

Answer:

H0: μ = 30; HA: μ ≠ 30

Step-by-step explanation:

1) Data given and notation    

$\bar X=38$ represent the mean of one way conmute time repellent for the sample    

$s=8$ represent the sample standard deviation

$n=100$ sample size  

$\mu_o =30$ represent the value that we want to test  

$\alpha$ not given 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)

2) State the null and alternative hypotheses.    

We need to conduct a hypothesis in order to determineif the average commute time of elementary school students in this district is different than the commonly cited standard of 30 minutes, the system of hypothesis would be:    

Null hypothesis:$\mu = 30$    

Alternative hypothesis:$\mu \neq 30$    

We don't know the population deviation, so for this case 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".

3) Calculate the statistic    

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

$t=\frac{38-30}{\frac{8}{\sqrt{100}}}=10$    

4) Calculate the P-value    

First we need to calculate the degrees of freedom given by:

$df=n-1=100-1=99$

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

$p_v =2*P(t_{99}>10)=0.000$

In Excel we can use the following formula to find the p value "=2*(1-T.DIST(10,99,TRUE))"  

5) Conclusion    

If we compare the p value and if we use a significance level, for example $\alpha=0.05$ we see that $p_v<\alpha$ so we can conclude that we reject the null hypothesis, so we can conclude that the average commute time of elementary school students in this district is different than the commonly cited standard of 30 minutes at 0.05 of signficance.