People suffering from hypertension, heart disease, or kidney problems may need t

Question

People suffering from hypertension, heart disease, or kidney problems may need to limit their intakes of sodium. The 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 Connecticut, for example, the notification level is 28 mg/L (milligrams per liter). Suppose that over the course of a particular year the mean concentration of sodium in the drinking water of a water system in Connecticut is 25.8 mg/L, and the standard deviation is 6 mg/L. Imagine that the water department selects a random sample of 31 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 31 specimens. If the mean exceeds 28 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. Even though the actual concentration of sodium in the drinking water is within the limit, there is a probability that the water department will erroneously advise its customers of an above-limit concentration of sodium. Suppose that the water department is willing to accept (at most) a 1% risk of erroneously notifying its customers that the sodium concentration is above the limit. A primary cause of sodium in the water supply is the salt that is applied to roadways during the winter to melt snow and ice. If the water department can't control the use of road salt and can't change the mean or the standard deviation of the sodium concentration in the drinking water, is there anything the department can do to reduce the risk of an erroneous notification to 1%?

Explanation / Answer

3.

a)

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    28      
u = mean =    25.8      
n = sample size =    31      
s = standard deviation =    6      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    2.0415136      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   2.0415136   ) =    0.020599902 [ANSWER, FIRST BLANK]

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b)

For a right tailed area of 0.01, the z score is

z = 2.326347874

Hence, as

z = (x-u)*sqrt(n)/sigma

Then

n = [z*sigma/(x-u)]^2 = (2.326347874*6/(28-25.8))^2 = 40.25376023 = 40 [ANSWER, C]

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