people are suffering from hypertension, heart diseases or kidney may need to lim

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people are suffering from hypertension, heart diseases or kidney may need to limit their intake of sodium. the public health department in some us states raded Assignment| Read Chapter 7 I Back to Assignment Due Tuesday 10.03.17 at 11:15 P Attempts: Average: 2 7. An application of the distribution of sample means Aa Aa 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 drínking water exceeds a designated limit. In Ontario, for example, the notification level is 20 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 Ontario is 18.3 mg/L, and the standard deviation is 6.mg/L Imagine that the water department selects a simple 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 20 mg/L, the water department notifies the public and recommends that people who are on sodi drinking water um restricted diets inform their physicians of the sodium content in their pe here to

Explanation / Answer

In Ontario, specification limit = 20 mg/ml

Let say X is the sodium concentration in drinking water.

Mean concentration of drinking water in Ontarion ? = 18.3 mg/l

Standard deviation of drinking water in Ontario ? = 6 mg/L

Sample size n = 31

so We have to find the probability that the sample mean of sample size 31 will not cross the specification limit .

Here,

Standard error of the sample se0  = ?/ sqrt(n) = 6/ sqrt(31) = 1.0776

Expected value of sample mean = population mean = 18.3 mg/l

so probability that sample mean will exceed the limit is

Pr(xbar > 20 mg/l; 18.3; 1.08) by normal distribution

Z = (20 - 18.3)/ 1.08 = 1.57

so Pr( Z > = 1.57) by standard norma distribution table

Pr( Z > = 1.57) = 1- 0.9418 = 0.0582

now that the probaiblity of crossing specification limit shall be reduced to 1% only so critical value of Z for the probability 0f 0.01 or less is

Z = 2.33

so let say the new sample size is n

Pr(Xbar >20.0 mg/ml ;18.3 ; 6/ sqrt(n)] > 2.33

(20 - 18.3)/ 6/ sqrt(n) > 2.33

1.7/ [6/ sqrt(n)] > 2.33

sqrt(n) > 8.2235

n = 68

Option D is correct. we shall increase the sample size to 68.

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