Question

In the US, fluoride is added to drinking water. In 2015, the US Department of Health [12] lowered the recommended amount of fluoride in drinking water. Fluoride is added to drinking water to help prevent tooth decay, but too much fluoride can lead to fluorosis (white stains in the enamel of the teeth). A particular city wants to check whether their drinking water meets the specifications made by the department of health. The department of health recommends that there should be 0.7 milligrams of fluoride per liter of water. Suppose we collect a random sample of 40 different liters of water and find the the average amount of fluoride per liter of water is 0.775 milligrams with standard deviation 0.2 milligrams. Conduct a hypothesis test using α = .01. Make sure to state the null and alternative hypotheses, state the appropriate test statistic, set up the rejection region, find the P-value for this test, make your conclusion using both the rejection region method and the P-value, and state your conclusion in the context of the problem. Also, find a 99% confidence interval for the true average amount of fluoride per liter of water and use this interval to make a conclusion about the test of hypothesis.

Answer 1

a)

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: μ = 0.7
Alternative Hypothesis, Ha: μ ≠ 0.7

Rejection Region
This is two tailed test, for α = 0.01 and df = 39
Critical value of t are -2.708 and 2.708.
Hence reject H0 if t < -2.708 or t > 2.708

Test statistic,
t = (xbar - mu)/(s/sqrt(n))
t = (0.775 - 0.7)/(0.2/sqrt(40))
t = 2.372

P-value Approach
P-value = 0.0227
As P-value >= 0.01, fail to reject null hypothesis.

Rejection Region Approach
As the value of test statistic, t is outside critical value range, fail to reject the null hypothesis

b)

sample mean, xbar = 0.775
sample standard deviation, s = 0.2
sample size, n = 40
degrees of freedom, df = n - 1 = 39

Given CI level is 99%, hence α = 1 - 0.99 = 0.01
α/2 = 0.01/2 = 0.005, tc = t(α/2, df) = 2.708

ME = tc * s/sqrt(n)
ME = 2.708 * 0.2/sqrt(40)
ME = 0.09

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (0.775 - 2.708 * 0.2/sqrt(40) , 0.775 + 2.708 * 0.2/sqrt(40))
CI = (0.69 , 0.86)

AS the interval contains 0.7 , fail to reject H0