Question

Unfortunately, arsenic occurs naturally in some ground water†. A mean arsenic level of μ = 8.0 parts per billion (ppb) is considered safe for agricultural use. A well in Texas is used to water cotton crops. This well is tested on a regular basis for arsenic. A random sample of 41 tests gave a sample mean of x = 7.2 ppb arsenic, with s = 2.1 ppb. Does this information indicate that the mean level of arsenic in this well is less than 8 ppb? Use α = 0.01.

Answer 1

`t=(7.2-8)/((2.1)/(√(41)))=-2.439`

`p_v =P(t_((40))<-2.439)=0.0096`

If we compare the p value and the significance level given
`\alpha=0.01` we see that
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, and the true mean is lower than 8 ppb at 1% of signficance. We need to be careful with this interpretation since the p value is near to the significance level.

Explanation:

Data given and notation

`\bar X=7.2` represent the mean height for the sample

`s=2.1/tex] represent the sample standard deviation </p><p>[tex]n=41` sample size

`\mu_o =8` 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.

We need to conduct a hypothesis in order to check if the true mean is less than 8, the system of hypothesis would be:

Null hypothesis:
`\mu \geq 8`

Alternative hypothesis:
`\mu < 8`

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=(\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=(7.2-8)/((2.1)/(√(41)))=-2.439`

P-value

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

`df=n-1=41-1=40`

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

`p_v =P(t_((40))<-2.439)=0.0096`

Conclusion

If we compare the p value and the significance level given
`\alpha=0.01` we see that
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, and the true mean is lower than 8 ppb at 1% of signficance. We need to be careful with this interpretation since the p value is near to the significance level.