/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 42 The Environmental Protection Age... [FREE SOLUTION] | 91Ó°ÊÓ

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The Environmental Protection Agency (EPA) recommends that the sodium content in public water supplies should be no more than \(20 \mathrm{mg}\) per liter (http://www.disabledworld.com/artman/publish/ sodiumwatersupply.shtml). Forty samples were taken from a large reservoir, and the amount of sodium in each sample was measured. The sample average was \(23.5 \mathrm{mg}\) per liter. Assume that the population standard deviation is \(5.6 \mathrm{mg}\) per liter. The EPA is interested in knowing whether the average sodium content for the entire reservoir exceeds the recommended level. If so, the communities served by the reservoir will have to be made aware of the violation. a. Find the \(p\) -value for the test of hypothesis. Based on this \(p\) -value, would the communities need to be informed of an excessive average sodium level if the maximum probability of a Type I error is to be \(.05 ?\) What if the maximum probability of a Type I error is to be \(.01 ?\) b. Test the hypothesis of part a using the critical-value approach and \(\alpha=.05 .\) Would the communities need to be notified? What if \(\alpha=.01 ?\) What if \(\alpha\) is zero?

Short Answer

Expert verified
The computations carried out in steps 1 to 5 lead to the conclusion about whether one should notify communities of an excess average sodium level. The actual notification decision depends on the result of hypothesis tests with different \(\alpha\) values.

Step by step solution

01

Calculate Test Statistic

The test statistic (z-score) for a one-sample mean with known population standard deviation is given by \(Z = \frac{\bar{X} - \mu}{\sigma / \sqrt{n}}\), where \(\bar{X}\) is the sample mean, \(\mu\) is the population mean under the null hypothesis, \(\sigma\) is the population standard deviation and n is the sample size. Substituting the given values, we get \(Z = \frac{23.5 - 20}{5.6 / \sqrt{40}}\).
02

Determine P-value

The P-value is the probability that we would observe a result as extreme or more extreme than the one we actually observed. We can look up the Z-value obtained from Step 1 in a standard normal (Z) distribution table or compute it using statistical software tools.
03

Decision for Type I error 0.05 and 0.01

We compare the P-value with the maximum probability of Type I error (0.05 and 0.01). If the P-value is less than or equal to the maximum probability of Type I error, we reject the null hypothesis.
04

Critical value approach for \(\alpha = 0.05\) and \(\alpha = 0.01\)

For this step, we find the critical value from the Z-distribution for the stated \(\alpha\) values. The decision rule states that we reject the null hypothesis if the test statistic is greater than the critical value.
05

Notification decision based on Hypothesis test

We decide on notifying communities if we end up rejecting the null hypothesis. Also, we discuss the implications of \(\alpha = 0\) in the context of the problem.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

z-score
The z-score is a statistical measurement that indicates where a particular data point falls in relation to the mean of a group of data points. It's especially useful in hypothesis testing to determine how far a sample mean is from the population mean under the null hypothesis. In this exercise, the formula for the z-score is \[ Z = \frac{\bar{X} - \mu}{\sigma / \sqrt{n}} \] where:
  • \( \bar{X} \) is the sample mean, which is 23.5 mg/L.
  • \( \mu \) is the hypothesized population mean, which is 20 mg/L.
  • \( \sigma \) is the population standard deviation, 5.6 mg/L.
  • \( n \) is the sample size, 40.
Substituting these values, we calculate the z-score to understand if the sample mean significantly deviates from the population mean, considering the variability measured by the standard deviation. Knowing the z-score helps us to look up or compute the probability of observing such a result, which directly aids in decision-making.
p-value
The p-value is a critical concept in hypothesis testing, representing the probability of obtaining a test statistic as extreme as one that was actually observed, assuming the null hypothesis is true. It quantifies the evidence against the null hypothesis. After calculating the z-score, we use it to find the corresponding p-value from the standard normal distribution. If this p-value is less than or equal to the significance level \( \alpha \), we reject the null hypothesis.For this exercise:
  • If the p-value is less than 0.05, the evidence is strong enough to conclude that the reservoir's sodium content may exceed the recommended level.
  • If it is less than 0.01, even stronger evidence suggests rejecting the null hypothesis.
By understanding p-values, we assess how "unexpected" our data is under the assumption that the null hypothesis is correct, aiding us in making informed decisions.
type I error
A Type I error occurs when we reject a true null hypothesis, essentially a false positive. In the context of the EPA's sodium test, such an error would mean alerting communities about a problem that doesn't exist. This can cause unnecessary worry and potentially costly corrective actions. The probability of making a Type I error is denoted by \( \alpha \), the significance level of the test. Common choices for \( \alpha \) are 0.05 and 0.01:
  • If \( \alpha = 0.05 \), there's a 5% risk of making a Type I error.
  • If \( \alpha = 0.01 \), the risk reduces to 1%.
Choosing \( \alpha \) involves a trade-off. A smaller \( \alpha \) reduces the false positive rate but may increase the chance of a Type II error (not rejecting a false null hypothesis). Understanding Type I errors helps in determining suitable \( \alpha \) levels for balancing risks and consequences.
critical value approach
The critical value approach is another method to decide on the rejection of the null hypothesis. In hypothesis testing, this involves comparing the calculated test statistic with critical values determined from the statistical distribution.In this problem:
  • For \( \alpha = 0.05 \), the critical z-value is approximately 1.645 for a one-tailed test.
  • For \( \alpha = 0.01 \), it's approximately 2.33.
  • If \( \alpha \) is zero, theoretically, any deviation would lead to rejecting the null hypothesis, though this is impractical.
If the test statistic exceeds the critical value, we reject the null hypothesis, suggesting the reservoir's sodium content exceeds the permissible level. This approach provides a clear cut-off point for decision-making, complementing the p-value in testing hypothesis.

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Most popular questions from this chapter

Make the following tests of hypotheses. a. \(H_{0}: \mu=25, \quad H_{1}: \mu \neq 25, \quad n=81, \quad \bar{x}=28.5, \quad \sigma=3, \quad \alpha=.01\) b. \(H_{0}: \mu=12, \quad H_{1}: \mu<12, \quad n=45, \quad \bar{x}=11.25, \quad \sigma=4.5, \quad \alpha=.05\) c. \(H_{0}: \mu=40, \quad H_{1}: \mu>40, \quad n=100, \quad \bar{x}=47, \quad \sigma=7, \quad \alpha=.10\)

The past records of a supermarket show that its customers spend an average of \(\$ 95\) per visit at this store. Recently the management of the store initiated a promotional campaign according to which each customer receives points based on the total money spent at the store, and these points can be used to buy products at the store. The management expects that as a result of this campaign, the customers should be encouraged to spend more money at the store. To check whether this is true, the manager of the store took a sample of 14 customers who visited the store. The following data give the money (in dollars) spent by these customers at this supermarket during their visits. \(\begin{array}{rrrrrrr}109.15 & 136.01 & 107.02 & 116.15 & 101.53 & 109.29 & 110.79 \\ 94.83 & 100.91 & 97.94 & 104.30 & 83.54 & 67.59 & 120.44\end{array}\) Assume that the money spent by all customers at this supermarket has a normal distribution. Using the \(5 \%\) significance level, can you conclude that the mean amount of money spent by all customers at this supermarket after the campaign was started is more than \(\$ 95 ?\) (Hint: First calculate the sample mean and the sample standard deviation for these data using the formulas learned in Sections \(3.1 .1\) and \(3.2 .2\) of Chapter \(3 .\) Then make the test of hypothesis about \(\mu .\) )

The manufacturer of a certain brand of auto batteries claims that the mean life of these batteries is 45 months. A consumer protection agency that wants to check this claim took a random sample of 24 such batteries and found that the mean life for this sample is \(43.05\) months. The lives of all such batteries have a normal distribution with the population standard deviation of \(4.5\) months. a. Find the \(p\) -value for the test of hypothesis with the alternative hypothesis that the mean life of these batteries is less than 45 months. Will you reject the null hypothesis at \(\alpha=.025\) ? b. Test the hypothesis of part a using the critical-value approach and \(\alpha=.025\).

Consider \(H_{0}: \mu=80\) versus \(H_{1}: \mu \neq 80\) for a population that is normally distributed. a. A random sample of 25 observations taken from this population produced a sample mean of 77 and a standard deviation of 8 . Using \(\alpha=.01\), would you reject the null hypothesis? b. Another random sample of 25 observations taken from the same population produced a sample mean of 86 and a standard deviation of \(6 .\) Using \(\alpha=.01\), would you reject the null hypothesis?

A random sample of 200 observations produced a sample proportion equal to \(.60 .\) Find the critical and observed values of \(z\) for each of the following tests of hypotheses using \(\alpha=.01\). a. \(H_{0}: p=.63\) versus \(H_{1}: p<.63\) b. \(H_{0}: p=.63\) versus \(H_{1}: p \neq .63\)

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