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Pairs of \(P\)-values and significance levels, \(\alpha\), are given. For each pair, state whether the observed \(P\)-value would lead to rejection of \(H_{0}\) at the given significance level. a. \(P\)-value \(=.084, \alpha=.05\) b. \(P\)-value \(=.003, \alpha=.001\) c. \(P\)-value \(=.498, \alpha=.05\) d. \(P\)-value \(=.084, \alpha=.10\) e. \(P\)-value \(=.039, \alpha=.01\) f. \(P\)-value \(=.218, \alpha=.10\)

Short Answer

Expert verified
Reject \( H_0 \) for pair d; do not reject \( H_0 \) for others.

Step by step solution

01

Understanding Hypothesis Testing

In hypothesis testing, a null hypothesis, denoted as \( H_{0} \), is typically tested. We compare the observed \( P \)-value to a pre-determined significance level, \( \alpha \). If the \( P \)-value is less than or equal to \( \alpha \), we reject \( H_{0} \). Otherwise, we do not reject \( H_{0} \).
02

Comparison for Pair a

Given \( P \)-value = 0.084 and \( \alpha = 0.05 \). Since 0.084 is greater than 0.05, we do not reject \( H_{0} \) for this pair.
03

Comparison for Pair b

Given \( P \)-value = 0.003 and \( \alpha = 0.001 \). Since 0.003 is greater than 0.001, we do not reject \( H_{0} \) for this pair.
04

Comparison for Pair c

Given \( P \)-value = 0.498 and \( \alpha = 0.05 \). Since 0.498 is greater than 0.05, we do not reject \( H_{0} \) for this pair.
05

Comparison for Pair d

Given \( P \)-value = 0.084 and \( \alpha = 0.10 \). Since 0.084 is less than 0.10, we reject \( H_{0} \) for this pair.
06

Comparison for Pair e

Given \( P \)-value = 0.039 and \( \alpha = 0.01 \). Since 0.039 is greater than 0.01, we do not reject \( H_{0} \) for this pair.
07

Comparison for Pair f

Given \( P \)-value = 0.218 and \( \alpha = 0.10 \). Since 0.218 is greater than 0.10, we do not reject \( H_{0} \) for this pair.

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Key Concepts

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

P-value
The concept of a P-value in hypothesis testing is fundamental. It helps us understand how extreme the observed data is. In simple terms, the P-value tells us the probability of observing our data, or something more extreme, if the null hypothesis is true.

Imagine you have a coin, and you suspect it is biased. You decide to test it by flipping it 100 times. If the coin is fair, you'd expect to get 50 heads. But what if you get 55 heads? The P-value would help you figure out how unusual that outcome is, assuming the coin is fair.
  • If the P-value is small, it means the observed data is unlikely under the null hypothesis.
  • P-values are usually expressed as decimals, such as 0.05 or 0.01.
A key point to remember is that a P-value does not tell us the probability that the null hypothesis is true. Instead, it provides the probability of the data assuming the null hypothesis is true.
Significance Level
The significance level, often denoted by the Greek letter \(\alpha\), is a threshold set by the researcher before conducting an experiment. This threshold determines at what point we will reject the null hypothesis. It represents the probability of making a false positive error, also known as a Type I error.

Common significance levels are 0.05, 0.01, and 0.10. Setting a significance level is like deciding how strict you want to be with the evidence required to reject the null hypothesis.
  • A lower \(\alpha\) value means you require stronger evidence to reject the null hypothesis.
  • Choosing \(\alpha=0.05\) implies accepting a 5% risk of incorrectly rejecting a true null hypothesis.
Choosing the right significance level is crucial, as it reflects the balance between caution and sensitivity needed for the study. It also impacts the probability of detecting a true effect when it exists.
Null Hypothesis
The null hypothesis, denoted as \(H_0\), is a statement that usually claims there is no effect or no difference. For example, if we hypothesize that a new drug has no effect on a disease, \(H_0\) would state precisely that.

The null hypothesis is the default assumption that we test against. The goal of hypothesis testing is to determine whether the data provides sufficient evidence to reject \(H_0\).
  • If you reject \(H_0\), it suggests that your data provides sufficient evidence against the null hypothesis.
  • If you do not reject \(H_0\), it suggests that there isn't enough evidence to support an alternative hypothesis.
A key point is that we never "accept" the null hypothesis; instead, we may not have enough evidence to reject it. Maintaining a neutral stance prevents incorrect conclusions that could arise from insufficient data.

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Most popular questions from this chapter

A random sample of 150 recent donations at a certain blood bank reveals that 82 were type A blood. Does this suggest that the actual percentage of type A donations differs from \(40 \%\), the percentage of the population having type A blood? Carry out a test of the appropriate hypotheses using a significance level of .01. Would your conclusion have been different if a significance level of 05 had heen nsed?

Let \(\mu\) denote the true average radioactivity level (picocuries per liter). The value \(5 \mathrm{pCi} / \mathrm{L}\) is considered the dividing line between safe and unsafe water. Would you recommend testing \(H_{0}: \mu=5\) versus \(H_{\mathrm{a}}: \mu>5\) or \(H_{0}: \mu=5\) versus \(H_{a}: \mu<5\) ? Explain your reasoning. [Hint: Think about the consequences of a type I and type II error for each possibility.]

To determine whether the pipe welds in a nuclear power plant meet specifications, a random sample of welds is selected, and tests are conducted on each weld in the sample. Weld strength is measured as the force required to break the weld. Suppose the specifications state that mean strength of welds should exceed \(100 \mathrm{Ib} / \mathrm{in}^{2}\), the inspection team decides to test \(H_{0}: \mu=100\) versus \(H_{a}: \mu>100\). Explain why it might be preferable to use this \(H_{\mathrm{a}}\) rather than \(\mu<100\).

Polymer composite materials have gained popularity because they have high strength to weight ratios and are relatively easy and inexpensive to manufacture. However, their nondegradable nature has prompted development of environmentally friendly composites using natural materials. The article "Properties of Waste Silk Short Fiber/Cellulose Green Composite Films" ( \(J\). of Composite Materials, 2012: 123-127) reported that for a sample of 10 specimens with \(2 \%\) fiber content, the sample mean tensile strength (MPa) was \(51.3\) and the sample standard deviation was 1.2. Suppose the true average strength for \(0 \%\) fibers (pure cellulose) is known to be \(48 \mathrm{MPa}\). Does the data provide compelling evidence for concluding that true average strength for the WSF/cellulose composite exceeds this value?

A sample of \(n\) sludge specimens is selected and the \(\mathrm{pH}\) of each one is determined. The one-sample \(t\) test will then be used to see if there is compelling evidence for concluding that true average \(\mathrm{pH}\) is less than 7.0. What conclusion is appropriate in each of the following situations? a. \(n=6, t=-2.3, \alpha=.05\) b. \(n=15, t=-3.1, \alpha=.01\) c. \(n=12, t=-1.3, \alpha=.05\) d. \(n=6, t=.7, \alpha=.05\) e. \(n=6, \bar{x}=6.68, s / \sqrt{n}=.0820\)

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