An application of the sampling distribution of the sample mean People suffering

Question

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 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 Massachusetts, 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 Massachusetts is 18 mg/L, and the standard deviation is 6 mg/L. Imagine that the water department selects a simple 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 necessary. 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%? No, there is nothing it can do. It can increase its sample size to n = 48. It can increase its sample size to n = 33. It can increase its sample size to n = 49.

Explanation / Answer

Solution:-

= 20, = 6, xbar = 18, n = 32

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: < 20

Alternative hypothesis: > 20

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the sample mean is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample t-test.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = s / sqrt(n)

S.E = 1.061

DF = n - 1 = 32 - 1

D.F = 31

t = (x - ) / SE

t = - 1.885

where s is the standard deviation of the sample, x is the sample mean, is the hypothesized population mean, and

The observed sample mean produced a t statistic test statistic of -1.885. We use the t Distribution Calculator to find P(t < - 1.885) = 0.0294

Interpret results. Since the P-value (0.0294) is less than the significance level (0.05), we have to reject the null hypothesis.

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