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A hypothesis will be used to test that a population mean equals 5 against the alternative that the population mean is less than 5 with known variance \(\sigma\). What is the criti- cal value for the test statistic \(Z_{0}\) for the following significance levels? (a) \(\alpha=0.01\) and \(n=20\) (b) \(\alpha=0.05\) and \(n=12\) (c) \(\alpha=0.10\) and \(n=15\)

Short Answer

Expert verified
(a) Z-critical value is -2.33, (b) -1.645, (c) -1.28.

Step by step solution

01

Understand the Null and Alternative Hypothesis

The null hypothesis is \( H_0: \mu = 5 \) while the alternative hypothesis is \( H_a: \mu < 5 \). This indicates a one-tailed (left-tailed) test in the Z-distribution.
02

Find the Z-Critical Value for Alpha 0.01

For a significance level, \( \alpha = 0.01 \), we look for a critical value in the standard normal distribution table corresponding to the cumulative probability of \( 0.01 \). The Z-critical value is approximately \( -2.33 \).
03

Find the Z-Critical Value for Alpha 0.05

For a significance level, \( \alpha = 0.05 \), we need the Z-critical value where the cumulative probability is \( 0.05 \). The Z-critical value here is approximately \( -1.645 \).
04

Find the Z-Critical Value for Alpha 0.10

For a significance level, \( \alpha = 0.10 \), find the Z-critical value where the cumulative probability is \( 0.10 \). The Z-critical value is about \( -1.28 \).

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Key Concepts

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

Z-distribution
In hypothesis testing, the Z-distribution is a fundamental concept. It refers to the normal distribution used when the sample size is large or the population variance is known. When we say something follows a Z-distribution, we mean that it has a mean of 0 and a standard deviation of 1. This is often referred to as the "standard normal distribution." Instead of working with various different normal distributions, statisticians standardize to this distribution, as it simplifies calculations. Another way to understand this is using a standardized test statistic, often denoted as "Z," which transforms any normal distribution into the standard normal distribution. By doing so, you can directly compare test statistics to critical values to understand significance. When dealing with a known variance, the Z-distribution helps us determine where our sample mean lies in relation to the population mean.
Critical Value
The critical value is an essential part of hypothesis testing. It helps you decide whether to reject the null hypothesis. Essentially, it serves as a cutoff point. If your test statistic falls beyond this point, then you have the evidence to reject the null hypothesis in favor of the alternative hypothesis. For a Z-test, critical values are determined using the Z-distribution table. These values change depending on the significance level, which reflects the probability threshold for rejecting the null hypothesis. For instance, in the original exercise:
  • with a significance level of 0.01, the critical value is approximately -2.33
  • for a significance level of 0.05, the critical value is about -1.645
  • for 0.10, it is around -1.28
By comparing your calculated Z-statistic to these critical values, you determine the strength of your results.
Significance Level
The significance level, denoted by \( \alpha \), plays a crucial role in hypothesis testing. It represents the probability of rejecting the null hypothesis when it is actually true. Commonly used significance levels are 0.01, 0.05, and 0.10.Choosing a significance level involves a trade-off. A lower significance level means you require stronger evidence against the null hypothesis before you reject it. For example, \( \alpha = 0.01 \) is stricter than \( \alpha = 0.05 \), as you are only willing to accept a 1% chance of wrongly rejecting the null.This choice directly influences the critical value. A smaller \( \alpha \) leads to a more negative critical value in left-tailed tests, thus requiring more evidence for rejection. Understanding significance levels is key to interpreting the results and making informed conclusions in hypothesis testing.
Population Mean
The population mean, often denoted by \( \mu \), is the average of all measurements in a particular population. In hypothesis testing, especially with Z-tests, you often want to see how the sample data compares to this mean.In the original exercise, the null hypothesis asserts that the population mean is 5. The alternative hypothesis suggests it is less than 5. By determining how the sample mean compares to 5 using the Z-test, you can assess whether the hypothesized population mean holds true.It's important to note that when conducting such tests, knowing the population variance or standard deviation aids in the application of the Z-test. This knowledge allows for determining the likelihood that a sample mean comes from a population with that specific mean. Accurately understanding and calculating around the population mean is fundamental to successful hypothesis testing.

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Most popular questions from this chapter

The advertised claim for batteries for cell phones is set at 48 operating hours, with proper charging procedures. A study of 5000 batteries is carried out and 15 stop operating prior to 48 hours. Do these experimental results support the claim that less than 0.2 percent of the company's batteries will fail during the advertised time period, with proper charging procedures? Use a hypothesis-testing procedure with \(\alpha=0.01\)

Supercavitation is a propulsion technology for undersea vehicles that can greatly increase their speed. It occurs above approximately 50 meters per second, when pressure drops sufficiently to allow the water to dissociate into water vapor, forming a gas bubble behind the vehicle. When the gas bubble completely encloses the vehicle, supercavitation is said to occur. Eight tests were conducted on a scale model of an undersea vehicle in a towing basin with the average observed speed \(\bar{x}=102.2\) meters per second. Assume that speed is normally distributed with known standard deviation \(\sigma=4\) meters per second. (a) Test the hypothesis \(H_{0}: \mu=100\) versus \(H_{1}: \mu<100\) using $$ \alpha=0.05 $$ (b) What is the \(P\) -value for the test in part (a)? (c) Compute the power of the test if the true mean speed is as low as 95 meters per second. (d) What sample size would be required to detect a true mean speed as low as 95 meters per second if we wanted the power of the test to be at least \(0.85 ?\) (e) Explain how the question in part (a) could be answered by constructing a one-sided confidence bound on the mean speed.

In each of the following situations, state whether it is a correctly stated hypothesis testing problem and why. (a) \(H_{0}: \mu=25, H_{1}: \mu \neq 25\) (b) \(H_{0}: \sigma>10, H_{1}: \sigma=10\) (c) \(H_{0}: \bar{x}=50, H_{1}: \bar{x} \neq 50\) (d) \(H_{0}: p=0.1, H_{1}: p=0.5\) (e) \(H_{0}: s=30, H_{1}: s>30\)

Suppose we are testing \(H_{0}: p=0.5\) versus \(H_{0}: p \neq 0.5 .\) Suppose that \(p\) is the true value of the population proportion. (a) Using \(\alpha=0.05,\) find the power of the test for \(n=100\), 150, and 300 assuming that \(p=0.6\). Comment on the effect of sample size on the power of the test. (b) Using \(\alpha=0.01\), find the power of the test for \(n=100\), 150 , and 300 assuming that \(p=0.6\). Compare your answers to those from part (a) and comment on the effect of \(\alpha\) on the power of the test for different sample sizes. (c) Using \(\alpha=0.05\), find the power of the test for \(n=100\), assuming \(p=0.08\). Compare your answer to part (a) and comment on the effect of the true value of \(p\) on the power of the test for the same sample size and \(\alpha\) level. (d) Using \(\alpha=0.01\), what sample size is required if \(p=0.6\) and we want \(\beta=0.05 ?\) What sample is required if \(p=0.8\) and we want \(\beta=0.05\) ? Compare the two sample sizes and comment on the effect of the true value of \(p\) on sample size required when \(\beta\) is held approximately constant.

The mean pull-off force of an adhesive used in marufacturing a connector for an automotive engine application should be at least 75 pounds. This adhesive will be used unless there is strong evidence that the pull-off force does not meet this requirement. A test of an appropriate hypothesis is to be conducted with sample size \(n=10\) and \(\alpha=0.05 .\) Assume that the pull-off force is normally distributed, and \(\sigma\) is not known. (a) If the true standard deviation is \(\sigma=1\), what is the risk that the adhesive will be judged acceptable when the true mean pull-off force is only 73 pounds? Only 72 pounds? (b) What sample size is required to give a \(90 \%\) chance of detecting that the true mean is only 72 pounds when \(\sigma=1 ?\) (c) Rework parts (a) and (b) assuming that \(\sigma=2\). How much impact does increasing the value of \(\sigma\) have on the answers you obtain?

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