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The desired percentage of silicon dioxide in a certain type of cement is \(5.0 \%\). A random sample of \(n=36\) specimens gave a sample average percentage of \(\bar{x}=5.21\). Let \(\mu\) be the true average percentage of silicon dioxide in this type of cement, and suppose that \(\sigma\) is known to be 0.38. Test \(H_{0}: \mu=5\) versus \(H_{a}: \mu \neq 5\) using a significance level of .01.

Short Answer

Expert verified
In Step 2, compute the value of the test statistic for the detailed decision in step 4. To summarize the final conclusion, it is necessary to complete the computation in step 2 and compare it with the critical values determined in step 3.

Step by step solution

01

Determine the necessary parameters

For this test, we have a sample with size \( n=36 \), the sample mean \( \overline{x}=5.21 \), population standard deviation \( \sigma=0.38 \), and the population mean under the null hypothesis \( \mu_{0}=5 \).
02

Calculate the test statistic

In order to calculate the test statistic, we use the formula: \( Z = \frac{\overline{x}-\mu_{0}}{\frac{\sigma}{\sqrt{n}}}\). After plug in the numbers, we have \( Z=\frac{5.21-5}{\frac{0.38}{\sqrt{36}}} \). Compute this to get the value of Z.
03

Determine the critical value

Given the significance level is 0.01, the critical values for a two-tailed test are \( Z_{\alpha/2}=-2.5758 \) and \( Z_{\alpha/2}=2.5758 \) from standard normal distribution table.
04

Make the decision

Compare the critical values with the test statistic. If test statistic falls in the acceptance region (which is between -2.5758 and 2.5758), then we fail to reject the null hypothesis. Otherwise, we reject null hypothesis in favor of the alternatives.

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Key Concepts

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

Z-test
A Z-test is a type of hypothesis test that determines if there's a significant difference between a sample mean and a population mean. It's particularly useful when the population variance is known, and the sample size is large (usually over 30). This statistical method employs the standard normal distribution (Z-distribution), which is why it is called a Z-test. In the exercise, the Z-test helps us determine whether the true average percentage of silicon dioxide in a cement sample deviates from the hypothesized value of 5%.To perform a Z-test:
  • Identify the sample mean (\(ar{x}\)) and the population mean under the null hypothesis (\(u_0\)).
  • Use the known population standard deviation (\(\sigma\)) and compute the standard error of the mean using the formula: \(\frac{\sigma}{\sqrt{n}}\).
  • Calculate the Z-statistic with the formula: \(Z = \frac{\bar{x}-u_0}{\frac{\sigma}{\sqrt{n}}}\).
In the provided exercise, we used a sample size of 36 with a sample mean of 5.21. The population standard deviation was noted as 0.38. By inserting these values into the Z-formula, we compute the Z-statistic to compare against critical values from the Z-distribution. This helps us decide whether to reject or retain our null hypothesis.
Significance Level
The significance level, commonly denoted as \(\alpha\), is a threshold that defines the probability of rejecting the null hypothesis when it is actually true. It essentially sets the bar for how extreme the test results must be in order to safely reject the null hypothesis. In hypothesis testing, this is also known as the type I error rate, representing the chance of making a false positive decision.

For example, a significance level of 0.01 means that there's a 1% risk of rejecting the null hypothesis when it is indeed true. In simple terms, we are 99% confident that our decision to reject or accept the null hypothesis is correct if we follow this threshold.

In the given exercise, a significance level of 0.01 has been utilized, meaning that the critical values from the Z-distribution are based on this extreme criterion. It impacts our acceptance region, where we only reject the null hypothesis if the Z-score lies beyond \(-2.5758\) to \(2.5758\). This rigorous threshold ensures a highly accurate test, minimizing the chances of erroneously claiming that the average percentage of silicon dioxide differs from 5%.
Central Limit Theorem
The Central Limit Theorem (CLT) is a fundamental principle in statistics. It explains why we can make inferences about population parameters from sample data. According to the CLT, the distribution of sample means will tend to follow a normal distribution (bell-shaped curve) as the sample size increases, regardless of the shape of the population distribution.This theorem makes it possible to apply normal probability theory to situations involving large samples, as it legitimizes the use of Z-tests when the sample size is sufficiently large (generally \(n > 30\)).

In the context of the provided exercise, the sample size was 36, which comfortably meets the requirement for applying the CLT. This allows us to assume that the sampling distribution of the sample mean is approximately normal, even if the original distribution of silicon dioxide percentages isn't. Consequently, the Z-test becomes applicable, and the results based on it are likely reliable.The CLT is especially powerful because it provides a link between intuitive, real-world observations and the mathematical frameworks of probability and statistics. It reassures us that as long as our samples are large and random, statistical inferences are trustworthy across various scenarios.To summarize:
  • The CLT allows the use of normal distribution for inference about sample means.
  • Requires a sufficiently large sample size, typically greater than 30.
  • Helps in justifying methods like the Z-test, ensuring accuracy in conclusions.

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Most popular questions from this chapter

The true average diameter of ball bearings of a certain type is supposed to be \(0.5\) in. What conclusion is appropriate when testing \(H_{0}: \mu=0.5\) versus \(H_{a}: \mu \neq 0.5 \mathrm{in}\) each of the following situations: a. \(n=13, t=1.6, \alpha=.05\) b. \(n=13, t=-1.6, \alpha=.05\) c. \(n=25, t=-2.6, \alpha=.01\) d. \(n=25, t=-3.6\)

A manufacturer of hand-held calculators receives large shipments of printed circuits from a supplier. It is too costly and time-consuming to inspect all incoming circuits, so when each shipment arrives, a sample is selected for inspection. Information from the sample is then used to test \(H_{0}: \pi=.05\) versus \(H_{a}: \pi>.05\), where \(\pi\) is the true proportion of defective circuits in the shipment. If the null hypothesis is not rejected, the shipment is accepted, and the circuits are used in the production of calculators. If the null hypothesis is rejected, the entire shipment is returned to the supplier because of inferior quality. (A shipment is defined to be of inferior quality if it contains more than \(5 \%\) defective circuits.) a. In this context, define Type I and Type II errors. b. From the calculator manufacturer's point of view, which type of error is considered more serious? c. From the printed circuit supplier's point of view, which type of error is considered more serious?

When a published article reports the results of many hypothesis tests, the \(P\) -values are not usually given. Instead, the following type of coding scheme is frequently used: \(^{*} p<.05,^{* *} p<.01,^{*} *^{*} p<.001, *\) Which of the symbols would be used to code for each of the following \(P\) -values? a. 037 c. \(.072\) b. \(.0026\) d. \(.0003\)

The report "2005 Electronic Monitoring \& Surveillance Survey: Many Companies Monitoring, Recording, Videotaping-and Firing-Employees" (American Management Association, 2005 ) summarized the results of a survey of 526 U.S. businesses. Four hundred of these companies indicated that they monitor employees' web site visits. For purposes of this exercise, assume that it is reasonable to regard this sample as representative of businesses in the United States. a. Is there sufficient evidence to conclude that more than \(75 \%\) of U.S. businesses monitor employees' web site visits? Test the appropriate hypotheses using a significance level of \(.01 .\) b. Is there sufficient evidence to conclude that a majority of U.S. businesses monitor employees' web site visits? Test the appropriate hypotheses using a significance level of \(.01\).

Water samples are taken from water used for cooling as it is being discharged from a power plant into a river. It has been determined that as long as the mean temperature of the discharged water is at most \(150^{\circ} \mathrm{F}\), there will be no negative effects on the river's ecosystem. To investigate whether the plant is in compliance with regulations that prohibit a mean discharge water temperature above \(150^{\circ} \mathrm{F}\), researchers will take 50 water samples at randomly selected times and record the temperature of each sample. The resulting data will be used to test the hypotheses \(H_{0}: \mu=150^{\circ} \mathrm{F}\) versus \(H_{a}: \mu>150^{\circ} \mathrm{F}\). In the context of this example, describe Type I and Type II errors. Which type of error would you consider more serious? Explain.

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