Review #2 Graded Assignment | Back to Assignment Due Sunday 01.21.18 at 1145 PM

Question

Review #2 Graded Assignment | Back to Assignment Due Sunday 01.21.18 at 1145 PM Attempts: 1 1. An application of the sampling distribution of the sample mean People suffering from hypertension, heart disease, or kidney problems may need to limit their intakes of sodium. The Keep the Highest: 1/2 public health departments in some U.S. states and Canadian provinces require community water systems to notity their customers if the sodium concentration in the drinking 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 18.6 mg/L, and the standard deviation is 6 mg/L Imagine that the water department selects a random sample of 32 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 32 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 questions, adjusting the parameters as r Normal Distribution Mean 19.5 Standard Deviation.0.85 12 14 16 18 20222 10 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.

Explanation / Answer

18.6 sd = 6 , n = 32

P(Xbar > 20)

P(t > (20 - 18.6 )/(6/sqrt(32))

= P(t> 1.31993)

= 1 - 0.9017 = 0.0983

hence 0.0934 is correct

b)

P(Xbar > 20) = 0.01

P(z > (20 - 18.6 )/(6/sqrt(n)) = 0.01

P(Z > z*) = 0.01

z* = 2.327

(20 - 18.6 )/(6/sqrt(n)) = 2.327

n is approx 100

option B) is correct

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