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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. Clin. Nutrit., 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
We do not reject the null hypothesis; there is no significant difference.

Step by step solution

01

Understanding Hypotheses

We are working with hypotheses where the null hypothesis is \( H_0: \mu = 5.63 \), which suggests that the average serum receptor concentration for pregnant women is the same as for all women, which is 5.63. The alternative hypothesis is \( H_a: \mu eq 5.63 \), implying that the average concentration for pregnant women differs from 5.63.
02

Considering the P-value

The problem states that the \( P \)-value is greater than 0.10. A \( P \)-value represents the probability of obtaining test results at least as extreme as the observed data under the null hypothesis. Since the \( P \)-value > 0.10 is much larger than the significance level of 0.01, this suggests there is weak evidence against the null hypothesis.
03

Significance Level Check

The significance level, \( \alpha \), is 0.01. This means we need a \( P \)-value less than 0.01 to reject the null hypothesis. Here, our \( P \)-value is greater than 0.10, which is much larger than 0.01, indicating that we lack evidence to reject the null hypothesis.
04

Conclusion

Since the \( P \)-value is greater than the significance level of 0.01, we do not have sufficient evidence to reject the null hypothesis \( H_0: \mu = 5.63 \). Therefore, we conclude that there is no significant difference in average serum receptor concentration between pregnant women and all women.

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Key Concepts

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

P-value interpretation
The P-value is a crucial concept in hypothesis testing. It helps us determine whether the observed data lead us to reject the null hypothesis. In simple terms, the P-value measures the likelihood of obtaining results at least as extreme as what was actually observed, assuming the null hypothesis is true.
When interpreting a P-value:
  • A small P-value (typically ≤ 0.05) suggests that the observed data are inconsistent with the null hypothesis, providing evidence to reject it.
  • A large P-value (> 0.05) indicates that the observed data are consistent with the null hypothesis, suggesting no strong evidence against it.
In this exercise, the P-value is reported as greater than 0.10. Since the P-value exceeds the common threshold (such as 0.05 or a more stringent level like 0.01), it indicates that the evidence is weak against the null hypothesis. Thus, we conclude that the data does not provide enough reason to reject the null hypothesis.
Significance level
The significance level, often denoted by \(\alpha\), is a threshold that determines when we can reject the null hypothesis. It reflects the maximum likelihood we are willing to accept of mistakenly rejecting a true null hypothesis, also known as a Type I error.
Common significance levels include:
  • 0.05 (5%) - A typical standard in many fields.
  • 0.01 (1%) - More conservative, used when more stringent testing is required.
In this problem, a significance level of 0.01 is used. This means we need very strong evidence – a P-value less than 0.01 – to reject the null hypothesis. Our actual P-value being greater than 0.10 does not meet this requirement. Thus, the test does not provide sufficient evidence to choose the alternative hypothesis over the null hypothesis.
Null and alternative hypotheses
In hypothesis testing, formulating null and alternative hypotheses is a foundational step. The null hypothesis, denoted as \(H_0\), represents the status quo or a statement of no effect or difference. It is the hypothesis that the test seeks to challenge.
Contrastingly, the alternative hypothesis, \(H_a\), is what you might want to prove. It suggests that there is a difference or effect.
For the exercise at hand:
  • The null hypothesis \(H_0\): \(\mu = 5.63\) posits that the average serum receptor concentration for pregnant women is the same as for all women.
  • The alternative hypothesis \(H_a\): \(\mu eq 5.63\) suggests that there is a measurable difference in this average.
Testing aims to gather evidence about these hypotheses using data. A test seeks to find sufficient reason to reject \(H_0\) in favor of \(H_a\). However, if the evidence is not strong enough, as in this case where the P-value is high, we continue to support the null hypothesis.

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Most popular questions from this chapter

Reconsider the paint-drying problem discussed in Example 9.2. The hypotheses were \(H_{0}: \mu=75\) versus \(H_{\mathrm{a}}: \mu<75\), with \(\sigma\) assumed to have value 9.0. Consider the alternative value \(\mu=74\), which in the context of the problem would presumably not be a practically significant departure from \(H_{0}\). a. For a level .01 test, compute \(\beta\) at this alternative for sample sizes \(n=100,900\), and 2500 . b. If the observed value of \(\bar{X}\) is \(\bar{x}=74\), what can you say about the resulting \(P\)-value when \(n=2500 ?\) Is the data statistically significant at any of the standard values of \(\alpha\) ? c. Would you really want to use a sample size of 2500 along with a level \(.01\) test (disregarding the cost of such an experiment)? Explain.

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The article "Statistical Evidence of Discrimination" (J. Amer. Statist. Assoc., 1982: 773-783) discusses the court case Swain v. Alabama (1965), in which it was alleged that there was discrimination against blacks in grand jury selection. Census data suggested that \(25 \%\) of those eligible for grand jury service were black, yet a random sample of 1050 people called to appear for possible duty yielded only 177 blacks. Using a level \(.01\) test, does this data argue strongly for a conclusion of discrimination?

Many older homes have electrical systems that use fuses rather than circuit breakers. A manufacturer of 40 -amp fuses wants to make sure that the mean amperage at which its fuses burn out is in fact 40 . If the mean amperage is lower than 40 , customers will complain because the fuses require replacement too often. If the mean amperage is higher than 40 , the manufacturer might be liable for damage to an electrical system due to fuse malfunction. To verify the amperage of the fuses, a sample of fuses is to be selected and inspected. If a hypothesis test were to be performed on the resulting data, what null and alternative hypotheses would be of interest to the manufacturer? Describe type I and type II errors in the context of this problem situation.

For healthy individuals the level of prothrombin in the blood is approximately normally distributed with mean \(20 \mathrm{mg} / 100 \mathrm{~mL}\) and standard deviation \(4 \mathrm{mg} / 100 \mathrm{~mL}\). Low levels indicate low clotting ability. In studying the effect of gallstones on prothrombin, the level of each patient in a sample is measured to see if there is a deficiency. Let \(\mu\) be the true average level of prothrombin for gallstone patients. a. What are the appropriate null and alternative hypotheses? b. Let \(\bar{X}\) denote the sample average level of prothrombin in a sample of \(n=20\) randomly selected gallstone patients. Consider the test procedure with test statistic \(\bar{X}\) and rejection region \(\bar{x} \leq 17.92\). What is the probability distribution of the test statistic when \(H_{0}\) is true? What is the probability of a type I error for the test procedure? c. What is the probability distribution of the test statistic when \(\mu=16.7\) ? Using the test procedure of part (b), what is the probability that gallstone patients will be judged not deficient in prothrombin, when in fact \(\mu=16.7\) (a type II error)? d. How would you change the test procedure of part (b) to obtain a test with significance level \(.05\) ? What impact would this change have on the error probability of part (c)? e. Consider the standardized test statistic \(Z=\) \((\bar{X}-20) /(\sigma / \sqrt{n})=(\bar{X}-20) / .8944\). What are the values of \(Z\) corresponding to the rejection region of part (b)?

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