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Chapter 8: Problem 32

A sample of 12 radon detectors of a certain type was selected, and each was exposed to \(100 \mathrm{pCi} / \mathrm{L}\) of radon. The resulting readings were as follows: $$ \begin{array}{rrrrrr} 105.6 & 90.9 & 91.2 & 96.9 & 96.5 & 91.3 \\ 100.1 & 105.0 & 99.6 & 107.7 & 103.3 & 92.4 \end{array} $$ a. Does this data suggest that the population mean reading under these conditions differs from 100 ? State and test the appropriate hypotheses using \(\alpha=.05\). b. Suppose that prior to the experiment, a value of \(\sigma=7.5\) had been assumed. How many determinations would then have been appropriate to obtain \(\beta=.10\) for the alternative \(\mu=95\) ?

Short Answer

Expert verified
a: Do not reject \( H_0 \), mean not different from 100. b: 60 determinations needed for power.

Step by step solution

01

State the Hypotheses

For part (a), we need to test if the population mean reading is different from 100.- Null Hypothesis (H_0): \( \mu = 100 \)- Alternative Hypothesis (H_a): \( \mu eq 100 \) This is a two-tailed test.
02

Calculate the Sample Mean and Standard Deviation

First, find the mean (\bar{x}) of the sample data: \[ \bar{x} = \frac{105.6 + 90.9 + 91.2 + 96.9 + 96.5 + 91.3 + 100.1 + 105.0 + 99.6 + 107.7 + 103.3 + 92.4}{12} \approx 98.7167 \]Next, calculate the sample standard deviation (s):\[ s = \sqrt{\frac{1}{n-1} \sum_{i=1}^{n} (x_i - \bar{x})^2} \approx 6.042 \]
03

Perform the Hypothesis Test

Use a t-test for the sample:The test statistic \( t \) is calculated as:\[ t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} = \frac{98.7167 - 100}{6.042/\sqrt{12}} \approx -0.678 \]Find the critical value for \( \alpha = 0.05 \) with \( n-1 = 11 \) degrees of freedom. In a two-tailed test, this corresponds to \( t_{0.025} \approx \pm2.201 \).Since \( -2.201 < -0.678 < 2.201 \), we do not reject \( H_0 \).
04

Determine Sample Size for Power

To find the sample size for part (b), use the formula for determining the sample size for a given power:\[ n = \left( \frac{(z_{\alpha/2} + z_{\beta}) \times \sigma}{\mu_0 - \mu_a} \right)^2 \]Given \( \sigma = 7.5 \), \( \alpha = 0.05 \), \( \beta = 0.10 \), \( \mu_0 = 100 \) and \( \mu_a = 95 \).- \( z_{0.025} = 1.96 \) and \( z_{0.1} = 1.28 \)Substitute into the formula:\[ n = \left( \frac{(1.96 + 1.28) \times 7.5}{100 - 95} \right)^2 = \left( \frac{3.24 \times 7.5}{5} \right)^2 \approx 59.29 \]Therefore, at least 60 determinations are required for the desired power.

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Key Concepts

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

t-test
The t-test is a statistical method used to determine if there is a significant difference between the means of two groups. In this context, it helps us decide if the average radon detector reading differs from 100. When performing a t-test, it's essential to have:
  • The sample mean, which is calculated from the data collected.
  • The sample standard deviation, which provides a measure of data variability.
  • The population mean, which is the hypothesized value we are comparing against.

In the exercise, we calculated the sample mean \( \bar{x} \approx 98.7167 \)and the sample standard deviation \( s \approx 6.042 \).

This information allows us to compute the t-statistic, which tells us how many standard deviations away our sample mean is from the hypothesized population mean.
sample size determination
Sample size determination is crucial for ensuring a study is adequately powered to detect an effect, if one exists. In this exercise, it addresses how many samples are needed to confidently detect if the mean radon reading is 95 instead of 100, given certain conditions. A proper sample size helps:
  • Increase the study's reliability, ensuring results are accurate.
  • Avoid Type II errors, which occur when we mistakenly accept a false null hypothesis.

The formula used combines the desired power (\( \beta \)) and the known standard deviation (\( \sigma \)) to compute the necessary sample size. Applying this formula, it's shown that at least 60 samples are essential for achieving the desired power when testing for a mean of 95.
two-tailed test
A two-tailed test is employed to determine if a sample mean is significantly different from a hypothesized mean, in either direction. Unlike a one-tailed test, which tests for deviations in only one direction, a two-tailed test checks both above and below the mean. This is why we calculate two critical values, representing both extremes of the distribution. The test is applicable when:
  • We are not specifically looking for an increase or decrease but any significant change.
  • The hypothesis is that the parameter could be greater or less than a particular value.

In our example, this approach is used to check if the mean detector reading differs from 100, not knowing whether higher or lower readings are expected. Since \( -2.201 < -0.678 < 2.201 \), we conclude that there's no significant difference.
statistical significance
Statistical significance offers a measure of whether our results are likely to be true for the entire population or if they occurred by chance. In this exercise, a significance level (\( \alpha \)) of 0.05 is used, meaning we accept a 5% risk of concluding that there is an effect when there is none.

Statistical significance relies on:
  • The calculated p-value or test statistic, which is compared against a critical value.
  • The assumption that results within the set threshold of significance can be generalized to the broader population.

For the detector readings, the test results were not statistically significant, indicating no evidence to conclude a notable difference in mean readings under the given study conditions.

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Most popular questions from this chapter

Water samples are taken from water used for cooling as it is being discharged from a power plant into a river. It has been determined that as long as the mean temperature of the discharged water is at most \(150^{\circ} \mathrm{F}\), there will be no negative effects on the river's ecosystem. To investigate whether the plant is in compliance with regulations that prohibit a mean discharge-water temperature above \(150^{\circ}, 50\) water samples will be taken at randomly selected times, and the temperature of each sample recorded. The resulting data will be used to test the hypotheses \(H_{0}: \mu=150^{\circ}\) versus \(H_{\mathrm{a}}: \mu>150^{\circ}\). In the context of this situation, describe type I and type II errors. Which type of error would you consider more serious? Explain.

After a period of apprenticeship, an organization gives an exam that must be passed to be eligible for membership. Let \(p=P\) (randomly chosen apprentice passes). The organization wishes an exam that most but not all should be able to pass, so it decides that \(p=.90\) is desirable. For a particular exam, the relevant hypotheses are \(H_{0}: p=.90\) versus the alternative \(H_{\mathrm{a}}: p \neq 90\). Suppose ten people take the exam, and let \(X=\) the number who pass. a. Does the lower-tailed region \(\\{0,1, \ldots, 5\\}\) specify a level \(.01\) test? b. Show that even though \(H_{\mathrm{a}}\) is two-sided, no two-tailed test is a level \(.01\) test. c. Sketch a graph of \(\beta\left(p^{\prime}\right)\) as a function of \(p^{\prime}\) for this test. Is this desirable?

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?

To obtain information on the corrosion-resistance properties of a certain type of steel conduit, 45 specimens are buried in soil for a 2-year period. The maximum penetration (in mils) for each specimen is then measured, yielding a sample average penetration of \(\bar{x}=52.7\) and a sample standard deviation of \(s=4.8\). The conduits were manufactured with the specification that true average penetration be at most 50 mils. They will be used unless it can be demonstrated conclusively that the specification has not been met. What would you conclude?

The sample average unrestrained compressive strength for 45 specimens of a particular type of brick was computed to be \(3107 \mathrm{psi}\), and the sample standard deviation was 188 . The distribution of unrestrained compressive strength may be somewhat skewed. Does the data strongly indicate that the true average unrestrained compressive strength is less than the design value of 3200 ? Test using \(\alpha=.001\).
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