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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. \(J\). 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; no evidence \(\mu \neq 5.63\).

Step by step solution

01

Understand the Hypotheses

The null hypothesis, denoted as \( H_0 \), is \( \mu = 5.63 \). This means that the true average serum receptor concentration for pregnant women is equal to the general average for all women. The alternative hypothesis, \( H_a \), is \( \mu eq 5.63 \), indicating that there is a difference.
02

Analyze the Given P-value

The problem states that the \( P \)-value for the test is greater than 0.10. Since \( P \)-value represents the probability of obtaining test results at least as extreme as the observed data, a \( P \)-value higher than 0.10 indicates weak evidence against the null hypothesis.
03

Compare P-value with Significance Level

We compare the \( P \)-value with the significance level \( \alpha = 0.01 \). A result is significant if the \( P \)-value is less than or equal to the significance level. Here, the \( P \)-value, which is greater than 0.10, is much larger than the significance level of 0.01.
04

Make a Decision

Since the \( P \)-value is greater than 0.01, we do not have enough evidence to reject the null hypothesis \( H_0: \mu = 5.63 \). Thus, we conclude that there is not enough statistical evidence to claim that the true average differs from 5.63.
05

State the Conclusion Clearly

At a significance level of 0.01, we fail to reject the null hypothesis. This means that the data does not provide sufficient evidence to suggest that the average serum receptor concentration for pregnant women is different from 5.63.

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Key Concepts

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

Null Hypothesis
The null hypothesis, often represented as \( H_0 \), is a foundational concept in hypothesis testing. It is a statement that assumes no effect or no difference in the context of a scientific inquiry. In the context of our exercise, the null hypothesis \( H_0: \mu = 5.63 \) suggests that the true average serum receptor concentration for pregnant women is equal to that of all women. It's essentially a claim of "no change" or "status quo."

When performing hypothesis testing, the goal is to assess the validity of the null hypothesis. We start by assuming \( H_0 \) is true, and then use statistical tests to determine whether the observed data provides strong enough evidence to reject this assumption. If the evidence is insufficient, we "fail to reject" the null hypothesis. However, failing to reject \( H_0 \) does not confirm it as true, but rather suggests that there isn't enough proof against it.

Understanding the null hypothesis helps in structuring experiments and framing what the outcomes might indicate. It provides a benchmark against which we measure the alternative hypothesis, denoted as \( H_a \), that suggests there is a meaningful difference or effect.
P-value
The \( P \)-value is a pivotal part of hypothesis testing. It quantifies the probability of obtaining a test statistic at least as extreme as the one observed, assuming that the null hypothesis \( H_0 \) is true. Essentially, \( P \)-value tells us how compatible the data is with \( H_0 \).

In our scenario, we have a \( P \)-value greater than 0.10, which indicates that the evidence against the null hypothesis is not strong. A high \( P \)-value suggests that the observed data could reasonably occur under the assumption that \( H_0 \) is true. Hence, in this context, there isn't sufficient statistical evidence to refute \( H_0 \).

When interpreting \( P \)-values, it's also crucial to avoid common misconceptions. A \( P \)-value does not provide the probability that the null hypothesis is true or false. Instead, it's about the evidence in the data. Misinterpreting what \( P \)-value represents could lead to incorrect conclusions in scientific research.
Significance Level
The significance level, denoted as \( \alpha \), is a threshold set by the researcher which determines when a result is considered statistically significant. In the exercise, a significance level of \( 0.01 \) is used, meaning there's a 1% risk of concluding that a difference exists when there is none. This level of significance represents a stringent criterion for decision making in hypothesis testing.

Comparison of the \( P \)-value to the significance level helps determine the outcome of the test. If the \( P \)-value is less than or equal to \( \alpha \), we reject the null hypothesis. However, if it is greater, as in our exercise where the \( P \)-value > 0.10, we fail to reject \( H_0 \), concluding that we don't have enough evidence to assert a difference at the 0.01 significance level.

Setting \( \alpha \) depends on the context of the study and the potential consequences of making a Type I error - mistakenly rejecting a true null hypothesis. A lower \( \alpha \) signifies a lower chance of achieving a false positive but requires stronger evidence against \( H_0 \) to declare a result significant.
Statistical Evidence
Statistical evidence refers to the support data provides for concluding about a hypothesis. It is crucial in making informed decisions in hypothesis testing. The quality of statistical evidence is assessed by how well it supports rejecting or failing to reject the null hypothesis.

In hypothesis testing, we rely heavily on statistical evidence translated through \( P \)-values and comparison to significance levels. In our example, the statistical evidence (\( P \)-value > 0.10) did not support rejecting the null hypothesis \( H_0: \mu = 5.63 \). Thus, it suggests that there is not a statistically significant difference between pregnant and non-pregnant women's serum receptor concentrations at the given significance level of 0.01.

However, it's important to remember that statistical evidence is only one part of decision-making in research. Practical significance and context should also be considered. While we might not find strong statistical evidence, the practical implications could still be vital for real-world applications.

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Most popular questions from this chapter

Automatic identification of the boundaries of significant structures within a medical image is an area of ongoing research. The paper "Automatic Segmentation of Medical Images Using Image Registration: Diagnostic and Simulation Applications" ( \(J\). of Medical Engr. and Tech., 2005: 53-63) discussed a new technique for such identification. A measure of the accuracy of the automatic region is the average linear displacement (ALD). The paper gave the following ALD observations for a sample of 49 kidneys (units of pixel dimensions). $$ \begin{array}{lllllll} 1.38 & 0.44 & 1.09 & 0.75 & 0.66 & 1.28 & 0.51 \\ 0.39 & 0.70 & 0.46 & 0.54 & 0.83 & 0.58 & 0.64 \\ 1.30 & 0.57 & 0.43 & 0.62 & 1.00 & 1.05 & 0.82 \\ 1.10 & 0.65 & 0.99 & 0.56 & 0.56 & 0.64 & 0.45 \\ 0.82 & 1.06 & 0.41 & 0.58 & 0.66 & 0.54 & 0.83 \\ 0.59 & 0.51 & 1.04 & 0.85 & 0.45 & 0.52 & 0.58 \\ 1.11 & 0.34 & 1.25 & 0.38 & 1.44 & 1.28 & 0.51 \end{array} $$ a. Summarize/describe the data. b. Is it plausible that ALD is at least approximately normally distributed? Must normality be assumed prior to calculating a CI for true average ALD or testing hypotheses about true average ALD? Explain. c. The authors commented that in most cases the ALD is better than or of the order of \(1.0\). Does the data in fact provide strong evidence for concluding that true average ALD under these circumstances is less than 1.0? Carry out an appropriate test of hypotheses. d. Calculate an upper confidence bound for true average ALD using a confidence level of \(95 \%\), and interpret this bound.

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