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The recommended daily dietary allowance for zinc among males older than age 50 years is \(15 \mathrm{mg} /\) day. The article "Nutrient Intakes and Dietary Patterns of Older Americans: A National Study" (J. of Gerontology, 1992: M145-150) reports the following summary data on intake for a sample of males age \(65-74\) years: \(n=115, \bar{x}=11.3\), and \(s=6.43\). Does this data indicate that average daily zinc intake in the population of all males ages \(65-74\) falls below the recommended allowance?

Short Answer

Expert verified
Yes, the average daily zinc intake is below 15 mg/day for males aged 65-74.

Step by step solution

01

Define Hypotheses

We need to determine whether the average zinc intake is statistically less than the recommended 15 mg/day. Therefore, we set up the null and alternative hypotheses as follows:\- Null Hypothesis (H0): \( \mu = 15 \) mg/day\- Alternative Hypothesis (H1): \( \mu < 15 \) mg/day. Here, \( \mu \) represents the true average zinc intake for the population of males aged 65-74.
02

Choose Significance Level

Select a significance level \( \alpha \) to test the hypothesis. Typically, \( \alpha = 0.05 \) is chosen.
03

Compute Test Statistic

Since the sample size \( n = 115 \) is large, we use the z-test for the population mean. Compute the test statistic using the formula:\[ z = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} \] where \( \bar{x} = 11.3 \), \( \mu = 15 \), \( s = 6.43 \), and \( n = 115 \). Substituting the values, we get:\[ z = \frac{11.3 - 15}{\frac{6.43}{\sqrt{115}}} \approx -6.12 \]
04

Determine Critical Value

For a one-tailed test at \( \alpha = 0.05 \), the critical value of the z-distribution is approximately -1.645. This value will be used to decide whether to reject the null hypothesis.
05

Make a Decision

Compare the computed test statistic \( z = -6.12 \) with the critical value of -1.645. Since \( -6.12 < -1.645 \), we reject the null hypothesis.
06

Conclude the Test

The rejection of the null hypothesis suggests that there is sufficient statistical evidence to conclude that the average daily zinc intake in the population of males aged 65-74 is indeed less than the recommended 15 mg/day.

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Key Concepts

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

Understanding Z-Tests
A z-test is a statistical method used when you want to test hypotheses about a population mean when the sample size is large and the population standard deviation is known. In the context of hypothesis testing, the z-test compares the sample mean to the population mean and checks how many standard deviations away the sample mean is from the hypothesized population mean.
The formula for a z-test is:
  • \( z = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} \)
Where:
  • \(\bar{x}\) is the sample mean
  • \(\mu\) is the population mean
  • \(s\) is the sample standard deviation
  • \(n\) is the sample size
It is essential to use a z-test when a sample size is large, like in our exercise with a sample of 115 males, as this helps achieve more reliable results.
The Null Hypothesis
In hypothesis testing, the null hypothesis, usually denoted as \(H_0\), is a statement that there is no effect or no difference, and it is the hypothesis that researchers typically try to disprove. In our zinc intake example, the null hypothesis is that the average zinc intake is equal to the recommended 15 mg/day. Symbolically, it is expressed as:
  • \(H_0: \mu = 15 \)
The objective is often to use data evidence to statistically reject \(H_0\), thereby allowing for the possibility that an alternative hypothesis (where a difference or effect exists) may be true. By rejecting the null hypothesis, conclusions can be drawn with a higher level of confidence.
Significance Level Explained
The significance level, denoted as \(\alpha\), determines the probability threshold at which you reject the null hypothesis. Common choices for \(\alpha\) include 0.05, 0.01, or 0.10, with 0.05 being most common. This means you are willing to accept a 5% chance of rejecting the null hypothesis when it is actually true.
The smaller the significance level, the stricter the test is for deciding whether to reject the null hypothesis. In our example, with \(\alpha = 0.05\), we are using a standard threshold, which balances between obtaining evidence against \(H_0\) and not rejecting it too easily. The choice of \(\alpha\) can affect the likelihood of Type I errors (falsely rejecting a true \(H_0\)).
Interpreting Critical Value
A critical value is the threshold that the test statistic must exceed in order to reject the null hypothesis. In a z-test, this value comes from a standard normal distribution table, determined by the chosen significance level.
For a one-tailed test at \(\alpha = 0.05\), a critical value is found at approximately -1.645 if you are testing for a value less than the hypothesized mean. This means if our calculated z-test statistic is less than -1.645, we reject \(H_0\).
In our zinc intake example, the test statistic calculated was -6.12, which is less than -1.645; therefore, it falls into the rejection region, which leads to rejecting the null hypothesis. Understanding such critical values helps us make informed decisions regarding our hypotheses.

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Most popular questions from this chapter

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?

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For which of the given \(P\)-values would the null hypothesis be rejected when performing a level .05 test? a. \(.001\) b. \(.021\) c. \(.078\) d. \(.047\) e. \(.148\)

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 \(=.218, \alpha=.10\)

One method for straightening wire before coiling it to make a spring is called "roller straightening." The article "The Effect of Roller and Spinner Wire Straightening on Coiling Performance and Wire Properties" (Springs, 1987: 27-28) reports on the tensile properties of wire. Suppose a sample of 16 wires is selected and each is tested to determine tensile strength \(\left(\mathrm{N} / \mathrm{mm}^{2}\right)\). The resulting sample mean and standard deviation are 2160 and 30, respectively. a. The mean tensile strength for springs made using spinner straightening is \(2150 \mathrm{~N} / \mathrm{mm}^{2}\). What hypotheses should be tested to determine whether the mean tensile strength for the roller method exceeds 2150 ? b. Assuming that the tensile strength distribution is approximately normal, what test statistic would you use to test the hypotheses in part (a)? c. What is the value of the test statistic for this data? d. What is the \(P\)-value for the value of the test statistic computed in part (c)? e. For a level \(.05\) test, what conclusion would you reach?

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