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

Question

People suffering from hypertension, heart disease, of 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 concentrating in the drinking water exceeds a designated limit in Ontario, for example, the notification level is 20 mg/L (m/grams 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 mg/L, and the standard deviating is 6 mg/L Imagine that the water department selects a simple random sample of 30 water specimens over the course of this year. Each spec men is sent to a lab for testing, and at the end of the year the water department computes the mean concentrating across the 30 specimens. If the mean exceeds 20 mq/L, the water department notices 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. Even though the actual concentrating 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 concentrating of sodium. Suppose that the water department is willing to accept (at most) a links of erroneously notifying its customers that the sodium concentrating 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 charge the mean or the standard deviating of the sodium concentrating in the drinking water, is there anything the department can do to reduce the ns' of an erroneous notification to 1%? It can increase its sample size to n = 34. It can increase its sample size to n = 48. No, there is netting it can do. It can increase its sample size to n = 49.

Explanation / Answer

In this case the data given to us is as follows:

Population mean m = 18

Population standard deviation, S = 6

Next we form a sampling distribution of sample means by taking many samples of size 30.

So sample size, n = 30

Using the Central Limit Theorem, we have the following parameters for the sampling distribution:

Mean, m = Population mean = 18

Standard deviation, also called as standard error, SE = S/(n^0.5) = 6/(30^0.5) = 1.095

Next, we pick a random sample from this distribution and calculate the probability that the sample mean is greater than 20.

So,

At X = 20, we have:

z = (X-m)/SE = (20-18)/1.095 = 1.826

So,

P(X > 20) = P(z > 1.826) = 0.034

So even though the actual sodium concentration is within the limit, there is a 0.034 probability that water department will erraneously devise customers of an above-limit concentration.

In this case we want a p-value of 0.01

So, the corresponding z-score is:

z = 2.33

So,

X = z*SE + m

Put the values:

20 = 2.33*SE + 18

SE = 0.858

Since SE = S/(n^0.5), so

S/(n^0.5) = 0.858

So,

6/(n^0.5) = 0.858

Solving we get:

n = 48.9 = 49

So the department can increase its sample size to 49.

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.