Question

The Department of Transportation of the State of New York claimed that it takes an average of 200 minutes to travel by train from New York to Buffalo. To test if the average travel time differs from 200 minutes, a random sample of 40 trains was taken and the average time required to travel from New York to Buffalo was 188 minutes, with a standard deviation of 28 minutes. What is the p-value for this test?

Answer 1

`t=(188-200)/((28)/(√(40)))=-2.7105`

`p_v =2*P(t_(39)<-2.7105)=0.004967`

Explanation:

Data given and notation

`\bar X=188` represent the sample mean

`s=28` represent the sample standard deviation

`n=40` sample size

`\mu_o =2-0` 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)

State the null and alternative hypotheses.

We need to apply a two tailed tailed test.

What are H0 and Ha for this study?

Null hypothesis:
`\mu = 200`

Alternative hypothesis :
`\mu \\eq 200`

Compute the test statistic

The statistic for this case is given by:

`t=(\bar X-\mu_o)/((s)/(√(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=(188-200)/((28)/(√(40)))=-2.7105`

Give the appropriate conclusion for the test

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

`df=n-1=40-1=39`

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

`p_v =2*P(t_(39)<-2.7105)=0.004967`

Conclusion

If we compare the p value and a significance level assumed for example
`\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 concldue that the true mean is significantly different from 200 minutes at 5% of significance.