Environmental Protection Agency standards require that the amount of lead in drinking water be less than 15 ppb. Twelve samples of water from a particular source have the following concentrations, in ppb. 11.4 13.9 11.2 14.5 15.2 8.1 12.4 8.6 10.5 17.1 9.8 15.9 A hypothesis test will be performed to determine whether the water from this source meets the EPA standard. Required:
a. State the appropriate null and alternate hypotheses.
b. Compute the P-value.
c. Can you conclude that the water from this source meets the EPA standard? Explain.

Using the t-distribution, we have that:

a)

The null hypothesis is:

The alternative hypothesis is:

b) The p-value is of 0.0051.

c) Since the p-value is of 0.0051, which is less than the standard significance level of 0.0051, it can be concluded that the mean is less than 15 ppb, and thus, this source meets the EPA standard.

Item a:

At the null hypothesis, it is tested if the mean is of at least 15 ppb, that is:

At the alternative hypothesis, it is tested if the mean is of less than 15 ppb, that is:

Item b:

We have the standard deviation for the sample, thus, the t-distribution is used. The test statistic is given by:

The parameters are:

In this problem, we have that . Additionally, using a calculator, the other parameters are:

Hence, the value of the test statistic is:

The p-value is found using a left-tailed test, as we are testing if the mean is less than a value, with t = -3.1 and 12 - 1 = 11 df.

Item c:

Since the p-value is of 0.0051, which is less than the standard significance level of 0.0051, it can be concluded that the mean is less than 15 ppb, and thus, this source meets the EPA standard.

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Step-by-step explanation:

Mean = (11.4 + 13.9 + 11.2 + 14.5 + 15.2 + 8.1 + 12.4 + 8.6 + 10.5 + 17.1 + 9.8 + 15.9)/12 = 12.4

Standard deviation = √(summation(x - mean)²/n

n = 12

Summation(x - mean)² = (11.4 - 12.4)^2 + (13.9 - 12.4)^2 + (11.2 - 12.4)^2+ (14.5 - 12.4)^2 + (15.2 - 12.4)^2 + (8.1 - 12.4)^2 + (12.4 - 12.4)^2 + (8.6 - 12.4)^2 + (10.5 - 12.4)^2 + (17.1 - 12.4)^2 + (9.8 - 12.4)^2 + (15.1 - 12.4)^2 = 89.62

Standard deviation = √(89.62/13) = 2.7

We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

a) For the null hypothesis,

µ ≤ 15

For the alternative hypothesis,

µ > 15

This is a right tailed test

b) Since the number of samples is small and no population standard deviation is given, the distribution is a student's t.

Since n = 12,

Degrees of freedom, df = n - 1 = 12 - 1 = 11

t = (x - µ)/(s/√n)

Where

x = sample mean = 12.4

µ = population mean = 15

s = samples standard deviation = 2.7

t = (12.4 - 15)/(2.7/√12) = - 3.34

We would determine the p value using the t test calculator. It becomes

p = 0.0034

c) Assuming level of significance = 0.05.

Since alpha, 0.05 > than the p value, 0.0034, then we would reject the null hypothesis. Therefore, At a 5% level of significance, we can conclude that the water from this source does meets the EPA standard. They are higher than 15ppb

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