Question

1. An application of the distribution of sample means Aa Aa People suffering from public health departments in some U.S. states and Canadian provinces require community water systems to notify their customers if the sodium concentration in the drinking water exceeds a designated limit. In Ontario, for exam hypertension, heart disease, or kidney problems may need to limit their intakes of sodium. The the notification level is 20 mg/L (milligrams per liter). Suppose that over the course of a particular year the mean concentration of sodium int system in Ontario is 18 mg/L, and the standard deviation is 6 mg/L he drinking water of a water Imagine that the water department selects a simple random sample of 30 water specimens over the course of this year. Each specimen is sent to a lab for testing, and at the end of the year the water department computes the mean concentration across the 30 specimens. If the mean exceeds 20 mg/L, the water department notifies the public and recommends that people who are on sodium-restricted diets inform their physicians of the sodium content in their drinking water. Use the Distributions tool to answer the following question. (Hint: Start by setting the mean and standard deviation parameters on the tool to the expected mean and standard error for the distribution of sample mean concentrations.)

Answer 1

The mean concentration of sodium in drinking water is 18 mg/L.

The upper limit of concentration of sodium in drinking water is marked 20 mg/L.

The sample size is 30, i.e. n=30.

The standard deviation is given 6mg/L.

H₀ : μ = 18 mg/L

H₀ : μ ≠ 18 mg/L

x̅ = 18 , σ = 6 , n = 30

Test statistic

z = (x̅−μ)(√n/σ) = (20-18)/(√30/6) = 1.82574

The tabulated value of z is  0.0339447

Thus, there is a  0.0339447 probability that the water department will erroneously advise its customers of an above limit concentration.

1% risk implies that the level of significancce (los) is 0.01.

At this los

$Z_{\alpha}=2.33$

Again,

z_α = (20−18)(√n/6) = 2.33

2√n = 13.98

√n = 6.99

n = 48.86

So,

n = 48.86 ~ 49

it can increase its sample size to 49