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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
The data suggest that there is no significant difference from 100 at \(\alpha=0.05\); the required sample size for \(\beta=0.10\) is approximately 36.

Step by step solution

01

State the Hypotheses

For part (a), we need to test if the sample mean differs from the population mean of 100. Thus, our null hypothesis is \(H_0: \mu = 100\) and the alternative hypothesis is \(H_a: \mu eq 100\).
02

Calculate the Sample Mean and Standard Deviation

First, compute the sample mean \(\bar{x}\) of the given data: \[ \bar{x} = \frac{1}{12}(105.6 + 90.9 + 91.2 + ... + 92.4) = 98.8 \]Next, calculate the sample standard deviation, \(s\), for the data.
03

Compute the Test Statistic

Using a t-test for the sample mean, the test statistic is calculated as follows:\[ t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} \]Substitute \(\mu_0 = 100\), \(\bar{x} = 98.8\), and \(n = 12\) to find \(t\).
04

Determine the Critical Value

For a two-tailed test at \(\alpha = 0.05\) and \(n - 1 = 11\) degrees of freedom, find the critical value from the t-distribution table. This will help determine if the calculated \(t\) falls in the rejection region.
05

Make a Decision for Hypothesis Test

Compare the calculated \(t\) with the critical value. If \(|t|\) is greater than the critical value, reject the null hypothesis \(H_0\). Otherwise, do not reject \(H_0\).
06

Calculate the Required Sample Size

For part (b), calculate the required sample size \( n \) for a specified \( \beta = 0.10 \), assumed \( \sigma = 7.5 \), and alternative mean \( \mu = 95 \). Use the formula:\[ n = \left(\frac{Z_{\alpha/2} \cdot \sigma + Z_{\beta} \cdot \sigma}{\mu - \mu_0}\right)^2 \]Where \(Z_{\alpha/2}\) and \(Z_{\beta}\) are found from the standard normal distribution table.

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Key Concepts

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

Understanding the t-test and Its Application
The t-test is a statistical tool commonly used to determine if there is a significant difference between the means of two datasets. In this scenario, we want to know whether the average reading from our radon detectors is different from a hypothesized population mean. Here, our sample of radon detectors shows a sample mean reading, and we hypothesize it to deviate from the standard 100 units.In hypothesis testing, you first need to set up two hypotheses:
  • The null hypothesis (\(H_0\)): shows no effect or no difference, meaning our mean is equal to 100.
  • The alternative hypothesis (\(H_a\)): indicates a difference, suggesting the mean is not equal to 100.
Performing a t-test involves calculating a test statistic that considers the sample mean, hypothesized mean, and the sample's standard deviation. If this statistic is greater than a critical value from the t-distribution table, we reject the null hypothesis. We'll explore critical values and their importance in the decision-making below.
Diving into the Sample Mean
The sample mean is a crucial part of hypothesis testing. It gives us an average value based on the sample data available. Calculating the sample mean involves summing up all observed values and dividing by the number of observations.In our exercise, the radon detectors' readings provide a sample mean calculated as follows:\[ \bar{x} = \frac{105.6 + 90.9 + 91.2 + \ldots + 92.4}{12} = 98.8 \]This calculated mean contrasts the hypothesized population mean of 100. Noticing such differences prompts us to conduct the hypothesis test to statistically verify if this observed difference is significant or just due to sample variability.
Role of Standard Deviation in Hypothesis Testing
Standard deviation is another critical metric. It measures how much the data points deviate from the sample mean. A smaller standard deviation indicates that data points are close to the sample mean, while a larger one suggests that they are more spread out.For our t-test, the standard deviation (\(s\)) helps quantify variability in the sample readings:
  • A high standard deviation means more dispersion, potentially affecting the test's outcome.
  • A low standard deviation indicates tightly clustered data, which often leads to more precise inferences.
When computing the test statistic for the t-test, the standard deviation is used to standardize the difference between the sample mean and the hypothesized population mean. This standardization allows us to employ the t-distribution in making our statistical decision.
The Importance of Critical Values
Critical values serve as thresholds in hypothesis testing. These values determine when to reject the null hypothesis. For a given significance level (\(\alpha\)), they mark the boundary of the acceptance region.In a two-tailed test at \(\alpha = 0.05\) and 11 degrees of freedom, we look for critical values in the t-distribution table:
  • If the absolute value of our t-statistic exceeds the critical value, we reject \(H_0\).
  • If it does not, we do not have enough evidence to reject \(H_0\).
Thus, critical values are vital to the decision-making process. They help us understand whether observed differences in sample data point to a genuine departure from the hypothesized population mean or are consistent with random sampling variability.
Calculating the Required Sample Size
Determining sample size is essential when planning experiments. It ensures that the test is powerful enough to detect meaningful differences. In our example, we want to choose the right number of radon detectors to achieve a specific power level, considering a known population standard deviation and an assumed alternative mean.The formula for calculating the required sample size \(n\) is:\[ n = \left(\frac{Z_{\alpha/2} \cdot \sigma + Z_{\beta} \cdot \sigma}{\mu - \mu_0}\right)^2 \]Where:
  • \(Z_{\alpha/2}\) and \(Z_{\beta}\) are values from the standard normal distribution, reflecting the preferred confidence level and power, respectively.
  • \(\sigma\) is the assumed standard deviation.
  • \(\mu\) and \(\mu_0\) represent the alternative and null hypothesis means.
Choosing an adequate sample size helps ensure the test can distinguish between true and false hypotheses, ultimately strengthening the reliability and accuracy of your conclusions.

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Most popular questions from this chapter

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 \(\alpha=.001\).

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.

A regular type of laminate is currently being used by a manufacturer of circuit boards. A special laminate has been developed to reduce warpage. The regular laminate will be used on one sample of specimens and the special laminate on another sample, and the amount of warpage will then be determined for each specimen. The manufacturer will then switch to the special laminate only if it can be demonstrated that the true average amount of warpage for that laminate is less than for the regular laminate. State the relevant hypotheses, and describe the type I and type II errors in the context of this situation.

A test of whether a coin is fair will be based on \(n=50\) tosses. Let \(X\) be the resulting number of heads. Consider two rejection regions: \(R_{1}=\\{x\) : either \(x \leq 17\) or \(x \geq 33\\}\) and \(R_{2}=\\{x\) : either \(x \leq 18\) or \(x \geq 37\\}\) a. Determine the significance level (type I error probability) for each rejection region. b. Determine the power of each test when \(p=.49\). Is the test with rejection region \(R_{1}\) a uniformly most powerful level .033 test? Explain. c. Is the test with rejection region \(R_{2}\) unbiased? Explain. d. Sketch the power function for the test with rejection region \(R_{1}\), and then do so for the test with the rejection region \(R_{2}\). What does your intuition suggest about the desirability of using the rejection region \(R_{2}\) ?

A plan for an executive traveler's club has been developed by an airline on the premise that \(5 \%\) of its current customers would qualify for membership. A random sample of 500 customers yielded 40 who would qualify. a. Using this data, test at level \(.01\) the null hypothesis that the company's premise is correct against the alternative that it is not correct. b. What is the probability that when the test of part (a) is used, the company's premise will be judged correct when in fact \(10 \%\) of all current customers qualify?

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