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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 critical value for the test statistic \(Z_{0}\) for the following significance levels? (a) 0.01 (b) 0.05 (c) 0.10

Short Answer

Expert verified
(a) \( Z_0 = -2.33 \), (b) \( Z_0 = -1.645 \), (c) \( Z_0 = -1.28 \)

Step by step solution

01

Understanding the Hypotheses

The null hypothesis is \( H_0: \mu = 5 \), and the alternative hypothesis is \( H_a: \mu < 5 \). This is a left-tailed test since the alternative hypothesis indicates that the population mean is less than 5.
02

Determine the Test Statistic

The test statistic used for this hypothesis test is the Z-statistic. We will find the critical value of \( Z_0 \) corresponding to each given significance level.
03

Critical Value for Significance Level 0.01

For a left-tailed test at a significance level of 0.01, look up the Z-table or use a calculator to find the Z-score that corresponds to a cumulative probability of 0.01. This value is approximately \( Z_0 = -2.33 \).
04

Critical Value for Significance Level 0.05

For a left-tailed test at a significance level of 0.05, look up the Z-table or use a calculator to find the Z-score that corresponds to a cumulative probability of 0.05. This value is approximately \( Z_0 = -1.645 \).
05

Critical Value for Significance Level 0.10

For a left-tailed test at a significance level of 0.10, look up the Z-table or use a calculator to find the Z-score that corresponds to a cumulative probability of 0.10. This value is approximately \( Z_0 = -1.28 \).

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Key Concepts

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

Population Mean
The population mean, often represented as \( \mu \), is a central concept in statistics. It is the average of all values in a given population. In hypothesis testing, we're typically interested in making inferences about this mean based on sample data. This mean tells us about the overall central tendency of a whole group of data and is a fundamental aspect of probability and statistical analysis.

In the context of hypothesis testing, we use the population mean to form hypotheses. For instance, when testing whether the population mean equals 5, our null hypothesis is stated as \( H_0: \mu = 5 \). Any deviation from this mean in our hypothesis testing would indicate whether or not we should accept or reject this hypothesis. The known variance \( \sigma^2 \) of the population also plays a crucial role in assessing how much the values are spread out around the mean. Together with the mean, the variance helps in determining the likelihood of the observed sample belonging to the population with the stated mean.
Significance Level
The significance level, symbolized by \( \alpha \), is a threshold in hypothesis testing. It's the probability of rejecting the null hypothesis when it is actually true, also known as the Type I error probability. This value is pre-determined by the researcher and often set at common levels such as 0.01, 0.05, or 0.10.

A lower significance level means you require stronger evidence against the null hypothesis to reject it. For instance, at \( \alpha = 0.01 \), there's only a 1% risk of claiming a statistically significant result when there isn't one, requiring more conservative evidence compared to a 10% risk at \( \alpha = 0.10 \). In our context, we are considering significance levels of 0.01, 0.05, and 0.10, each influencing the determination of our critical value.
Critical Value
A critical value helps determine the borderline between the acceptance region and the rejection region for the null hypothesis. In hypothesis testing, it is derived based on the chosen significance level. The critical value depends on the nature of the test. For a one-tailed test like the left-tailed tests we're considering, the critical value is identified from a standard normal distribution (Z-distribution).

To find the critical value for significance levels 0.01, 0.05, and 0.10, we use a Z-table or statistical software. For the scenarios described:
  • At \( \alpha = 0.01 \), the critical value is approximately \( Z_0 = -2.33 \).
  • At \( \alpha = 0.05 \), the critical value is approximately \( Z_0 = -1.645 \).
  • At \( \alpha = 0.10 \), the critical value is approximately \( Z_0 = -1.28 \).
These values tell us how extreme the test statistic must be in order to reject the null hypothesis at each significance level.
Z-statistic
The Z-statistic is a vital component of hypothesis testing when dealing with known population variance. It quantifies how many standard deviations an element is from the mean. In a hypothesis test, we compute the Z-statistic using the formula: \[Z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}}\]Where:
  • \( \bar{x} \) is the sample mean.
  • \( \mu \) is the population mean stated in the null hypothesis.
  • \( \sigma \) is the population standard deviation.
  • \( n \) is the sample size.

In testing whether a population mean is less than a specified value, we compare the computed Z-statistic against the critical value derived for our chosen significance level. A Z-statistic that falls below the critical value indicates sufficient evidence to reject the null hypothesis. This process ensures that we base our decision not on random chance but on the statistical analysis of our sample data.

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Most popular questions from this chapter

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 cooling system in a nuclear submarine consists of an assembly of welded pipes through which a coolant is circulated. Specifications require that weld strength must meet or exceed 150 psi. (a) Suppose that the design engineers decide to test the hypothesis \(H_{0}: \mu=150\) versus \(H_{1}: \mu>150 .\) Explain why this choice of alternative hypothesis is better than \(H_{1}: \mu<150\) (b) A random sample of 20 welds results in \(\bar{x}=153.7\) psi and \(s=11.3\) psi. What conclusions can you draw about the hypothesis in part (a)? State any necessary assumptions about the underlying distribution of the data.

When \(X_{1}, X_{2}, \ldots, X_{n}\) are independent Poisson random variables, each with parameter \(\lambda,\) and \(n\) is large, the sample mean \(\bar{X}\) has an approximate normal distribution with mean \(\lambda\) and variance \(\lambda / n\). Therefore, $$ Z=\frac{\bar{X}-\lambda}{\sqrt{\lambda / n}} $$ has approximately a standard normal distribution. Thus we can test \(H_{0}: \lambda=\lambda_{0}\) by replacing \(\lambda\) in \(Z\) by \(\lambda_{0}\). When \(X_{i}\) are Poisson variables, this test is preferable to the largesample test of Section \(9-2.3,\) which would use \(S / \sqrt{n}\) in the denominator, because it is designed just for the Poisson distribution. Suppose that the number of open circuits on a semiconductor wafer has a Poisson distribution. Test data for 500 wafers indicate a total of 1038 opens. Using \(\alpha=0.05,\) does this suggest that the mean number of open circuits per wafer exceeds \(2.0 ?\)

The heat evolved in calories per gram of a cement mixture is approximately normally distributed. The mean is thought to be 100 and the standard deviation is 2 . We wish to test \(H_{0}: \mu=100\) versus \(H_{1}: \mu \neq 100\) with a sample of \(n=9\) specimens. (a) If the acceptance region is defined as \(98.5 \leq \bar{x} \leq 101.5\), find the type I error probability \(\alpha\). (b) Find \(\beta\) for the case where the true mean heat evolved is 103 . (c) Find \(\beta\) for the case where the true mean heat evolved is 105. This value of \(\beta\) is smaller than the one found in part (b) above. Why?

For the hypothesis test \(H_{0}: \mu=5\) against \(H_{1}: \mu<5\) and variance known, calculate the \(P\) -value for each of the following test statistics. (a) \(z_{0}=2.05\) (b) \(z_{0}=-1.84\) (c) \(z_{0}=0.4\)

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