Question

A researcher in a local transportation department claims that the average number of cars crossing a bridge in town per day is greater than 500. She takes a sample of the last 30 days and finds and average of 520 with a standard deviation of 50.

1. With a .01 level of significance, is there enough evidence that the average number of the cars passing the bridge per day is greater than 500?

Answer 1

`p_v =P(t_(29)>2.191)=0.0183`

If we compare the p value and a significance level given
`\alpha=0.01` we see that
`p_v>\alpha` so we can conclude that we FAIL to reject the null hypothesis, and the actual true mean for the time is not significantly higher than 500 at 1% of significance.

Explanation:

Data given and notation

`\bar X=520` represent the sample mean

`s=50` represent the standard deviation for the sample

`n=30` sample size

`\mu_o =500` represent the value that we want to test

`\alpha=0.01` 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 to be tested

We need to conduct a hypothesis in order to determine if the mean number of cars crossing a bridge in town per day is greater than 500, the system of hypothesis would be:

Null hypothesis:
`\mu \leq 500`

Alternative hypothesis:
`\mu > 500`

Compute the test statistic

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=(\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".

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

`t=(520-500)/((50)/(√(30)))=2.191`

Now we need to find the degrees of freedom for the t distirbution given by:

`df=n-1=30-1=29`

What do you conclude?

a. Use the critical value approach.

Assuming 99% of confidence and
`\alpha=0.01` we can use the t distribution with 29 degrees of freedom in order to calculate a critical value that accumulates 0.01 of the area on the right tail of the distribution. We can use excel and the code to do this is given by: "=T.INV(1-0.01,29)". And we got the critical value
`t_(\alpha)=2.462`.

Since our calculated value < critical value. We fail to reject the null hypothesis, and we can say that at 1% of significance we don't have enough evidence to conclude that the true mean is higher than 500.

b. Use the p-value approach

Since is a one right tailed test the p value would be:

`p_v =P(t_(29)>2.191)=0.0183`

If we compare the p value and a significance level given
`\alpha=0.01` we see that
`p_v>\alpha` so we can conclude that we FAIL to reject the null hypothesis, and the actual true mean for the time is not significantly higher than 500 at 1% of significance.

Answer 2

We accept H₀ we dont have evidence to claim that the number of cars passing per day is bigger than 500

Explanation:

We assume Normal Distribution

Population mean μ₀

Population standard deviation σ = 50

Sample mean μ = 520

Population sample n n = 30

significance level α = 0,01

We are going to develop a right tail-test

1.- Hipothesis Test:

H₀ = null hypothesis μ₀ = 500

Hₐ = Alternative hypothesis μ₀ > 500

2.- Confidence interval 99 %

significance level α = 0,01

We have to interpolate

From z table we get:

0.01017 ⇒ 2.32

0.00990 ⇒ 2.33

Δ 0.00027 ⇒ 0.01

Then 0.00027 ⇒ 0.01 0.01017 - 0.01 = 0.00017

0.00017 ⇒ x ??

x = 0.00629 and 2.32 + 0.00629

z(c) = 2.326

3.-Compute z(s)

z(s) = [ μ - μ₀ ] / σ/√n ⇒ z(s) = 20 *√30 / 50

z(s) = 2.19

4.-Compare z(s) and z(c)

Z(c) > z(s) 2.326 > 2.19

So z(s) is inside acceptance region We accept H₀

We dont have evidence to believe than cars passing the bridge per day are greater than 500.

Note: If we think that n = 30 is not enough sample and went to s-student tables we will find t(c) = 2.4662 ( for α = 0.01 one tail test and 29 degree of fredom ) which will imply that again our z(s) is inside acceptance region)