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Chapter 9: Problem 49

A hypothesis will be used to test that a population mean equals 10 against the alternative that the population mean is greater than 10 with known variance \(\sigma\). What is the critical 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) 2.33, (b) 1.645, (c) 1.28

Step by step solution

01

Understanding the Hypothesis Test Type

This is a one-tailed hypothesis test since we are testing if the population mean is greater than 10. We will use the standard normal distribution to find the critical value for the test statistic because the population variance is known.
02

Determine the Significance Level

The significance level, denoted by \( \alpha \), determines the region in the tail for the critical value. The given \( \alpha \) values are 0.01, 0.05, and 0.10 for parts (a), (b), and (c) respectively.
03

Find Critical Value for \( \alpha = 0.01 \)

For significance level \( \alpha = 0.01 \), the critical value \( Z_{0} \) is found by looking it up in the standard normal distribution table. The critical value for a one-tailed test is the \( Z \)-score that leaves 1% in the upper tail: \( Z_{0} = 2.33 \).
04

Find Critical Value for \( \alpha = 0.05 \)

For significance level \( \alpha = 0.05 \), find the \( Z \)-score that leaves 5% in the upper tail. The critical value \( Z_{0} = 1.645 \).
05

Find Critical Value for \( \alpha = 0.10 \)

With \( \alpha = 0.10 \), locate the \( Z \)-score covering 10% in the upper tail. The critical value \( Z_{0} = 1.28 \).
06

Summarize the Critical Values

The critical values of the \( Z \)-statistic for the respective significance levels are: \( Z_{0} = 2.33 \) for \( \alpha = 0.01 \), \( Z_{0} = 1.645 \) for \( \alpha = 0.05 \), and \( Z_{0} = 1.28 \) for \( \alpha = 0.10 \). These values indicate where to reject the null hypothesis in the upper tail.

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Key Concepts

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

One-tailed Test
A one-tailed test is used in hypothesis testing when we want to determine if a parameter is either greater than or less than a certain value. Unlike a two-tailed test, which examines possibilities on both sides of the distribution, a one-tailed test focuses on one direction. In this case, we are testing whether the population mean is greater than 10.
It's like asking: "Is the average score higher than 10?" By doing this, we concentrate all of our significance level on one side of the distribution.
  • When to use: Choose a one-tailed test when your research hypothesis predicts a direction.
  • Examples: Testing if a new drug increases patient recovery rates.
Overall, using a one-tailed test can give us more power to detect an effect in the expected direction, which can be crucial when resources are limited or stakes are high.
Significance Level
The significance level, represented by \( \alpha \), is a threshold set by researchers to determine the likelihood of rejecting a true null hypothesis. It's a way of saying "how much chance of error am I willing to accept?"
In the context of the original problem, we were given \( \alpha \) values as 0.01, 0.05, and 0.10.
  • \( \alpha = 0.01 \): Strict criterion, allowing only a 1% chance of committing a Type I error.
  • \( \alpha = 0.05 \): Commonly used in research, balances risk and sensitivity.
  • \( \alpha = 0.10 \): A more lenient threshold, provides a balanced approach when pilot testing.
It's crucial to select an appropriate \( \alpha \) level based on the context and potential consequences of your decision. Researchers often default to 0.05, but in life-critical areas, you should consider a lower significance level to minimize errors.
Critical Values
Critical values mark the boundary where we decide whether to reject or fail to reject the null hypothesis. For a standard normal distribution, these values correspond to the \( Z \)-score.
In our exercise, the critical values for various significance levels were determined:
  • \( Z_0 = 2.33 \) for \( \alpha = 0.01 \)
  • \( Z_0 = 1.645 \) for \( \alpha = 0.05 \)
  • \( Z_0 = 1.28 \) for \( \alpha = 0.10 \)
To find these values, we look up a standard normal distribution table or use statistical software.
These critical values are pivotal because they tell us the cut-off points beyond which the null hypothesis should be rejected. In other words, if our test statistic falls beyond the critical threshold, we consider the results statistically significant.
Z-distribution
The Z-distribution, also known as the standard normal distribution, is a normal distribution with a mean of 0 and a standard deviation of 1. It serves as the backbone for many statistical tests, including our one-tailed tests.
In hypothesis testing, the test statistic calculated from the sample data is often represented as a \( Z \)-score when referring to this distribution.
  • Properties: Symmetrical, bell-shaped curve.
  • Usage: Facilitates comparison of different data sets by standardizing test scores.
This distribution helps us determine how far away our sample score is from the mean under the null hypothesis assumption. Using the Z-distribution allows us to make decisions about hypotheses irrespective of the original data's scale.

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Most popular questions from this chapter

The life in hours of a battery is known to be approximately normally distributed, with standard deviation \(\sigma=1.25\) hours. A random sample of 10 batteries has a mean life of \(\bar{x}=40.5\) hours. (a) Is there evidence to support the claim that battery life exceeds 40 hours? Use \(\alpha=0.05 .\) (b) What is the \(P\) -value for the test in part (a)? (c) What is the \(\beta\) -error for the test in part (a) if the true mean life is 42 hours? (d) What sample size would be required to ensure that \(\beta\) does not exceed 0.10 if the true mean life is 44 hours? (e) Explain how you could answer the question in part (a) by calculating an appropriate confidence bound on life.

Consider the computer output below. One-Sample T: Test of \(m u=100\) vs not \(=100\) $$ \begin{array}{llllclll} \text { Variable } & \mathrm{N} & \text { Mean } & \text { StDev } & \text { SE Mean } & 95 \% \mathrm{CI} & \mathrm{T} & \mathrm{P} \\ \mathrm{X} & 16 & 98.33 & 4.61 & ? & (?, ?) & ? & ? \end{array} $$ (a) How many degrees of freedom are there on the \(t\) -statistic? (b) Fill in the missing information. You may use bounds on the \(P\) -value. (c) What are your conclusions if \(\alpha=0.05 ?\) (d) What are your conclusions if the hypothesis is \(H_{0}: \mu=\) 100 versus \(H_{0}: \mu>100 ?\)

Output from a software package is given below: One-Sample Z: Test of \(m u=14.5 \mathrm{vs}>14.5\) The assumed standard deviation \(=1.1\) $$ \begin{array}{lccccccc} \text { Variable } & \mathrm{N} & \text { Mean } & \text { StDev } & \text { SE Mean } & \mathrm{Z} & \mathrm{P} \\ \mathrm{x} & 16 & 15.016 & 1.015 & ? & ? & ? \end{array} $$ (a) Fill in the missing items. What conclusions would you draw? (b) Is this a one-sided or a two-sided test? (c) Use the normal table and the above data to construct a \(95 \%\) lower bound on the mean. (d) What would the \(P\) -value be if the alternative hypothesis is \(H_{1}: \mu \neq 14.5 ?\)

For the hypothesis test \(H_{0}: \mu=10\) against \(H_{1}: \mu>10\) with variance unknown and \(n=15,\) approximate the \(P\) -value for each of the following test statistics. (a) \(t_{0}=2.05\) (b) \(t_{0}=-1.84\) (c) \(t_{0}=0.4\)

For the hypothesis test \(H_{0}: \mu=7\) against \(H_{1}: \mu \neq 7\) 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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