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Chapter 8: Problem 53

Let \(\mu\) denote true average serum receptor concentration for all pregnant women. The average for all women is known to be \(5.63\). The article "Serum Transferrin Receptor for the Detection of Iron Deficiency in Pregnancy" (Amer. \(J\). of Clinical Nutr:, 1991: 1077-1081) reports that \(P\)-value \(>.10\) for a test of \(H_{0}: \mu=5.63\) versus \(H_{\mathrm{a}}: \mu \neq 5.63\) based on \(n=176\) pregnant women. Using a significance level of \(.01\), what would you conclude?

Short Answer

Expert verified
Fail to reject the null hypothesis; insufficient evidence to say \( \mu \neq 5.63 \).

Step by step solution

01

Understand the hypothesis

The null hypothesis \( H_0 \) is that the true average serum receptor concentration \( \mu \) for pregnant women is equal to 5.63. The alternative hypothesis \( H_a \) is that \( \mu eq 5.63 \). We are given a \( P \)-value greater than 0.10 for the test and need to use a significance level of 0.01.
02

Compare P-value with significance level

The \( P \)-value given is greater than 0.10, which is much higher than our significance level of 0.01. Since \( P \)-value > 0.01, it means that the observed data is not statistically significant at the \( 0.01 \) level.
03

Make a conclusion about the null hypothesis

Since the \( P \)-value is greater than the significance level, we fail to reject the null hypothesis \( H_0 \). This means there is not enough evidence to conclude that the true average for pregnant women differs from 5.63.

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Key Concepts

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

P-value
In hypothesis testing, the P-value is a tool that helps us decide whether the evidence we have is strong enough to support a particular claim against a pre-stated assumption. Simply put, the P-value tells us how probable our observed data is, assuming that the null hypothesis is true.
  • A low P-value (typically less than the significance level) indicates that the observed data is unlikely under the null hypothesis, leading us to consider alternative possibilities.
  • A high P-value suggests that the observed data isn't unusual under the null hypothesis, implying that the null could be true.
For example, in our exercise, the P-value is greater than 0.10. This tells us there's a relatively high chance that the observed average serum receptor concentration could occur even if the hypothesis (\( H_0: \mu = 5.63 \)) is correct.
Significance Level
The significance level, often denoted as \( \alpha \), is a critical threshold in hypothesis testing. It helps us determine whether a P-value is sufficiently small to reject the null hypothesis. The significance level represents the probability of accidentally rejecting a true null hypothesis鈥攁 risk often called Type I error.
  • Common choices for significance levels are 0.01, 0.05, and 0.10, although the choice depends on the specific context and the consequences of making incorrect decisions.
  • In our exercise, the significance level is set at 0.01, making it a stringent criterion to ensure that we only reject the null hypothesis when there is strong evidence against it.
Given the P-value > 0.10, which is much higher than our chosen significance level of 0.01, we conclude that the result is not statistically significant enough to reject the null hypothesis.
Null Hypothesis
The null hypothesis, represented as \( H_0 \), serves as a starting assumption for statistical testing. In many cases, it posits that there is no effect or not a significant difference in the data being analyzed. By assessing evidence against \( H_0 \), we can make informed decisions about whether a new hypothesis (the alternative hypothesis) is more plausible.
  • For the exercise, \( H_0 \) claims that the average serum receptor concentration for pregnant women is 5.63, matching the general population average.
  • The alternative hypothesis \( H_a \) suggests that this average is different for pregnant women.
In the solution, because the P-value is more than 0.01, we "fail to reject" the null hypothesis. This isn鈥檛 a confirmation of \( H_0 \) being right, but it indicates insufficient evidence to take the stance of \( H_a \). Hence, the data doesn't favor a difference in averages for pregnant versus all women, as tested under this hypothesis.

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Most popular questions from this chapter

A sample of 50 lenses used in eyeglasses yields a sample mean thickness of \(3.05 \mathrm{~mm}\) and a sample standard deviation of \(.34 \mathrm{~mm}\). The desired true average thickness of such lenses is \(3.20 \mathrm{~mm}\). Does the data strongly suggest that the true average thickness of such lenses is something other than what is desired? Test using \(\alpha=.05\).

Let \(X_{1}, \ldots, X_{n}\) denote a random sample from a normal population distribution with a known value of \(\sigma\). a. For testing the hypotheses \(H_{0}: \mu=\mu_{0}\) versus \(H_{\mathrm{a}}: \mu>\mu_{0}\) (where \(\mu_{0}\) is a fixed number), show that the test with test statistic \(\bar{X}\) and rejection region \(\bar{x} \geq \mu_{0}+2.33 \sigma / \sqrt{n}\) has significance level \(.01\). b. Suppose the procedure of part (a) is used to test \(H_{0}: \mu \leq \mu_{0}\) versus \(H_{\mathrm{a}}: \mu>\mu_{0}\). If \(\mu_{0}=100, n=25\), and \(\sigma=5\), what is the probability of committing a type I error when \(\mu=99\) ? When \(\mu=98\) ? In general, what can be said about the probability of a type I error when the actual value of \(\mu\) is less than \(\mu_{0}\) ? Verify your assertion.

The article "Uncertainty Estimation in Railway Track LifeCycle Cost" (J. of Rail and Rapid Transit, 2009) presented the following data on time to repair (min) a rail break in the high rail on a curved track of a certain railway line. A normal probability plot of the data shows a reasonably linear pattern, so it is plausible that the population distribution of repair time is at least approximately normal. The sample mean and standard deviation are \(249.7\) and \(145.1\), respectively. a. Is there compelling evidence for concluding that true average repair time exceeds \(200 \mathrm{~min}\) ? Carry out a test of hypotheses using a significance level of \(.05\). b. Using \(\sigma=150\), what is the type II error probability of the test used in (a) when true average repair time is actually \(300 \mathrm{~min}\) ? That is, what is \(\beta(300)\) ?

A hot-tub manufacturer advertises that with its heating equipment, a temperature of \(100^{\circ} \mathrm{F}\) can be achieved in at most \(15 \mathrm{~min}\). A random sample of 42 tubs is selected, and the time necessary to achieve a \(100^{\circ} \mathrm{F}\) temperature is determined for each tub. The sample average time and sample standard deviation are \(16.5 \mathrm{~min}\) and \(2.2 \mathrm{~min}\), respectively. Does this data cast doubt on the company's claim? Compute the \(P\)-value and use it to reach a conclusion at level .05.

Before agreeing to purchase a large order of polyethylene sheaths for a particular type of high-pressure oil-filled submarine power cable, a company wants to see conclusive evidence that the true standard deviation of sheath thickness is less than \(.05 \mathrm{~mm}\). What hypotheses should be tested, and why? In this context, what are the type I and type II errors?
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