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Chapter 9: Problem 68

The error in a measurement is normally distributed with mean \(\mu\) and standard deviation 1 . Consider a random sample of \(n\) errors, and show that the likelihood ratio test for \(H_{0}: \mu=0\) versus \(H_{\mathrm{a}}: \mu \neq 0\) rejects the null hypothesis when either \(\bar{x} \geq c\) or \(\bar{x} \leq-c\). What is \(c\) for a test with \(\alpha=.05\) ? How does the test change if the standard deviation of an error is \(\sigma_{0}\) (known) and the relevant hypotheses are \(H_{0}: \mu=0\) versus \(H_{\mathrm{a}}: \mu \neq \mu_{0}\) ?

Short Answer

Expert verified
Reject \(H_0\) if \(|\bar{x}| \geq \frac{1.96}{\sqrt{n}}\). With \(\sigma_0\), use \(|\bar{x}| \geq \frac{1.96 \sigma_0}{\sqrt{n}}\).

Step by step solution

01

Understanding the Hypotheses and Model

We are testing the hypotheses: \(H_0: \mu = 0\) versus \(H_a: \mu eq 0\). The errors are normally distributed with a standard deviation of 1. The sample mean \(\bar{x}\) is used to perform the test with a sample size \(n\). Our goal is to determine when the test statistic indicates rejection of the null hypothesis.
02

Likelihood Ratio Test Statistic

The likelihood ratio test statistic for \(\mu\) under the normal distribution is based on \(Z = \bar{x} \sqrt{n}\). Under \(H_0: \mu = 0\), \(\bar{x}\) follows a normal distribution with mean 0 and standard deviation \(1/\sqrt{n}\), so \(Z\) follows a standard normal distribution, \(N(0, 1)\).
03

Determining Critical Value or Rejection Region

For a two-tailed test with \(\alpha = 0.05\), the critical values are given by the standard normal distribution as \(\pm z_{\alpha/2}\). Since \(\alpha = 0.05\), \(\alpha/2 = 0.025\), we find \(z_{0.025} = 1.96\). Thus, the rejection region is when \(|Z| = |\bar{x} \sqrt{n}| \geq 1.96\).
04

Relation to Sample Mean

Rewriting \(|Z| \geq 1.96\) in terms of \(\bar{x}\), this becomes \(|\bar{x}| \geq \frac{1.96}{\sqrt{n}}\). Let \(c = \frac{1.96}{\sqrt{n}}\). The null hypothesis \(H_0: \mu = 0\) is rejected if \(\bar{x} \geq c\) or \(\bar{x} \leq -c\).
05

Updating for General Standard Deviation \(\sigma_0\)

If the standard deviation is \(\sigma_0\), then the test statistic becomes \(Z = \frac{\bar{x}\sqrt{n}}{\sigma_0}\). Under \(H_0: \mu = 0\), \(Z\) is standard normal. The rejection region is then \(|\bar{x}| \geq \frac{1.96 \sigma_0}{\sqrt{n}}\). For \(H_a: \mu eq \mu_0\), the calculations for \(Z\) adjust accordingly but involve \(\mu_0\), impacting \(\bar{x}\)'s position relative to \(\mu_0\).

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Key Concepts

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

Normal Distribution
The normal distribution is one of the most fundamental concepts in statistics. It describes how values of a variable are distributed in a population. Often called a "bell curve" due to its characteristic shape, it represents how data points are spread out around the mean.
  • A normal distribution is symmetric, meaning the left and right sides of the graph are mirror images.
  • It's defined by two key parameters: the mean (\(\mu\)), which is the center of the distribution, and the standard deviation (\(\sigma\)), which measures the spread of data around the mean.
  • In the context of this exercise, the errors in measurement are normally distributed, which means the majority of errors will cluster around the mean value, \(\mu = 0\).
The property of this normal distribution allows us to use it in various statistical tests, like the likelihood ratio test. Since the distribution of measurement errors follows this pattern, determining probabilities and defining rejection regions for hypothesis tests becomes feasible.
Hypothesis Testing
Hypothesis testing is a statistical method that allows us to make inferences about a population parameter based on sample data. In the simplest form, it helps decide between two competing hypotheses:
  • The null hypothesis (\(H_0\)): It asserts a specific value or state about a population parameter. Here, \(\mu = 0\), suggesting no error in the measurements from the expected mean.
  • The alternative hypothesis (\(H_a\)): It proposes that the population parameter is different from the null. In this case, \(\mu eq 0\), indicating potential deviation from the expected mean.
The likelihood ratio test employed in the exercise helps determine how probable our sample data is under each hypothesis. By calculating a specific test statistic and comparing it to critical values derived from the normal distribution, we can ascertain whether to reject \(H_0\).
For the test with significance level \(\alpha = 0.05\), we seek critical values that define the rejection regions for \(H_0\). If our test statistic falls within these regions, we reject \(H_0\), making us more confident in \(H_a\). This approach helps in understanding not only the probable errors in measurement but also guides corrective actions.
Standard Deviation
Standard deviation is a measure that quantifies the amount of variation or dispersion in a set of values. In simpler terms, it tells us how much the individual data points deviate from the mean:
  • A smaller standard deviation indicates that the data points tend to be closer to the mean.
  • A larger standard deviation means the data points are spread out over a wider range of values.
In the exercise, the standard deviation plays a crucial role in determining the test's rejection region. Initially, it is given as 1, making calculations straightforward. The critical value \(c\) for rejection regions is found using \(c = \frac{1.96}{\sqrt{n}}\).
When the standard deviation is not 1 but another known value, \(\sigma_0\), the formula for the test statistic adapts accordingly. It becomes \(Z = \frac{\bar{x}\sqrt{n}}{\sigma_0}\), allowing for consistent hypothesis testing across varying conditions. Thus, the standard deviation directly affects the width of the confidence intervals and the sensitivity of the hypothesis test.

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Most popular questions from this chapter

When \(X_{1}, X_{2}, \ldots, X_{n}\) are independent Poisson variables, each with parameter \(\lambda\), and \(n\) is large, the sample mean \(\bar{X}\) has approximately a normal distribution with \(\mu=E(\bar{X})=\lambda\) and \(\sigma^{2}=V(\bar{X})=\) \(\lambda / n\). This implies that $$ Z=\frac{\bar{X}-\lambda}{\sqrt{\lambda / n}} $$ has approximately a standard normal distribution. For testing \(H_{0}: \lambda=\lambda_{0}\), we can replace \(\lambda\) by \(\lambda_{0}\) in the equation for \(Z\) to obtain a test statistic. This statistic is actually preferred to the large-sample statistic with denominator \(S / \sqrt{n}\) (when the \(X_{i}\) 's are Poisson) because it is tailored explicitly to the Poisson assumption. If the number of requests for consulting received by a certain statistician during a 5 -day work week has a Poisson distribution and the total number of consulting requests during a 36 -week period is 160 , does this suggest that the true average number of weekly requests exceeds 4.0? Test using \(\alpha=.02\).

The article "Caffeine Knowledge, Attitudes, and Consumption in Adult Women" \((J\). Nutrit. Ed., 1992: 179-184) reports the following summary data on daily caffeine consumption for a sample of adult women: \(n=47, \bar{x}=215 \mathrm{mg}, s=235\) \(\mathrm{mg}\), and range \(=5-1176\). a. Does it appear plausible that the population distribution of daily caffeine consumption is normal? Is it necessary to assume a normal population distribution to test hypotheses about the value of the population mean consumption? Explain your reasoning. b. Suppose it had previously been believed that mean consumption was at most \(200 \mathrm{mg}\). Does the given data contradict this prior belief? Test the appropriate hypotheses at significance level \(.10\) and include a \(P\)-value in your analysis.

The true average breaking strength of ceramic insulators of a certain type is supposed to be at least \(10 \mathrm{psi}\). They will be used for a particular application unless sample data indicates conclusively that this specification has not been met. A test of hypotheses using \(\alpha=.01\) is to be based on a random sample of ten insulators. Assume that the breaking-strength distribution is normal with unknown standard deviation. a. If the true standard deviation is \(.80\), how likely is it that insulators will be judged satisfactory when true average breaking strength is actually only 9.5? Only 9.0? b. What sample size would be necessary to have a \(75 \%\) chance of detecting that true average breaking strength is \(9.5\) when the true standard deviation is \(.80\) ?

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

Scientists have recently become concerned about the safety of Teflon cookware and various food containers because perfluorooctanoic acid (PFOA) is used in the manufacturing process. An article in the July 27, 2005, New York Times reported that of 600 children tested, \(96 \%\) had PFOA in their blood. According to the FDA, \(90 \%\) of all Americans have PFOA in their blood. a. Does the data on PFOA incidence among children suggest that the percentage of all children who have PFOA in their blood exceeds the FDA percentage for all Americans? Carry out an appropriate test of hypotheses. b. If \(95 \%\) of all children have PFOA in their blood, how likely is it that the null hypothesis tested in (a) will be rejected when a significance level of \(.01\) is employed? c. Referring back to (b), what sample size would be necessary for the relevant probability to be \(.10\) ?
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