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Let the test statistic \(Z\) have a standard normal distribution when \(H_{0}\) is true. Give the significance level for each of the following situations: a. \(H_{\mathrm{a}}: \mu>\mu_{0}\), rejection region \(z \geq 1.88\) b. \(H_{\mathrm{a}}: \mu<\mu_{0}\), rejection region \(z \leq-2.75\) c. \(H_{\mathrm{a}}: \mu \neq \mu_{0}\), rejection region \(z \geq 2.88\) or \(z \leq-2.88\)

Short Answer

Expert verified
a = 0.0301, b = 0.0030, c = 0.0040

Step by step solution

01

Understand the Standard Normal Distribution

The test statistic, denoted as \( Z \), follows a standard normal distribution, which has a mean of 0 and a standard deviation of 1. This means we use the standard normal table (or Z-table) to find the probabilities associated with specific Z-values.
02

Determine Significance Level for Case (a)

In case (a), the alternative hypothesis \( H_{a}: \mu > \mu_{0} \) implies a right-tailed test. The rejection region is \( z \geq 1.88 \). To find the significance level, calculate the probability that \( Z \geq 1.88 \). Using the Z-table, find the probability: \( P(Z < 1.88) = 0.9699 \). Hence, the significance level \( \alpha \) is equal to \( 1 - 0.9699 = 0.0301 \).
03

Determine Significance Level for Case (b)

For case (b), the hypothesis \( H_{a}: \mu < \mu_{0} \) denotes a left-tailed test, with the rejection region \( z \leq -2.75 \). Find the probability that \( Z \leq -2.75 \) using the Z-table: \( P(Z < -2.75) = 0.0030 \). Therefore, the significance level \( \alpha \) is 0.0030.
04

Determine Significance Level for Case (c)

In case (c), we have a two-tailed test since \( H_{a}: \mu eq \mu_{0} \), with rejection regions \( z \geq 2.88 \) or \( z \leq -2.88 \). Find the probabilities at both tails: \( P(Z \geq 2.88) = 1 - P(Z < 2.88) = 1 - 0.9980 = 0.0020 \) and \( P(Z \leq -2.88) = 0.0020 \). The total significance level is the sum of the probabilities in both tails, \( \alpha = 0.0020 + 0.0020 = 0.0040 \).

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Key Concepts

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

Standard Normal Distribution
The standard normal distribution is a fundamental concept in statistics. It is a special kind of normal distribution with a mean of 0 and a standard deviation of 1. This distribution is very useful because it allows comparisons across different tests and measurements. When we say that a test statistic, like a Z-score, follows a standard normal distribution, we mean that it can be compared against this curve. The Z-score itself measures how many standard deviations away a particular value is from the mean. In hypothesis testing, the standard normal distribution is used to determine the probability of observing a test statistic as extreme, or more extreme than, the observed value if the null hypothesis is true. This probability is crucial in deciding whether to reject or fail to reject the null hypothesis.
Significance Level
The significance level, denoted as \( \alpha \), plays a key role in hypothesis testing. It represents the probability of rejecting the null hypothesis when it is actually true, which is known as a Type I error. Commonly chosen significance levels are 0.05, 0.01, or 0.10, depending on the study's requirements. When planning an experiment, researchers decide on the significance level beforehand to ensure that the test's results are interpreted correctly. In the context of our exercise:
  • In case (a), the significance level is calculated by finding the probability that the test statistic is greater than or equal to 1.88, which was 0.0301.
  • In case (b), it was 0.0030, highlighting a much stricter test criterion.
  • In case (c), we dealt with a two-tailed test with a significance level of 0.0040.
Understanding and choosing the significance level is crucial as it informs how confident you can be in your test results.
Z-table
The Z-table, also known as the standard normal table, is a tool that helps you find the probability that a standard normal random variable is less than or equal to a given value. In simpler terms, it tells you how likely it is for a score to be below a certain point on the standard normal distribution curve. Using the Z-table involves looking up a Z-score on the table, which then gives you the cumulative probability up to that Z-score. This is particularly helpful when determining significance levels or critical values in hypothesis testing. For example, in step 2 of our solution:
  • We used the Z-table to find that \( P(Z < 1.88) = 0.9699 \).
  • This probability helped us calculate the significance level \( \alpha = 1 - 0.9699 = 0.0301 \).
By understanding how to use a Z-table, you can effectively find the necessary probabilities to conduct hypothesis tests.
Rejection Region
The concept of a rejection region is integral to hypothesis testing. It defines the range of values for which we reject the null hypothesis. If the test statistic falls into this region, the observed result is considered statistically significant. In the context of our problem:
  • For a right-tailed test like in case (a), the rejection region was defined as \( z \geq 1.88 \).
  • In a left-tailed test like case (b), the rejection region was \( z \leq -2.75 \).
  • A two-tailed test, like in case (c), involves rejection regions at both ends: \( z \geq 2.88 \) and \( z \leq -2.88 \).
The rejection region is determined based on the significance level. It helps decide whether the statistical evidence is strong enough to reject the null hypothesis.

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Most popular questions from this chapter

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?

A manufacturer of nickel-hydrogen batteries randomly selects 100 nickel plates for test cells, cycles them a specified number of times, and determines that 14 of the plates have blistered. a. Does this provide compelling evidence for concluding that more than \(10 \%\) of all plates blister under such circumstances? State and test the appropriate hypotheses using a significance level of 05 . In reaching your conclusion, what type of error might you have committed? b. If it is really the case that \(15 \%\) of all plates blister under these circumstances and a sample size of 100 is used, how likely is it that the null hypothesis of part (a) will not be rejected by the level .05 test? Answer this question for a sample size of 200 . c. How many plates would have to be tested to have \(\beta(.15)=.10\) for the test of part (a)?

Show that for any \(\Delta>0\), when the population distribution is normal and \(\sigma\) is known, the two-tailed test satisfies \(\beta\left(\mu_{0}-\Delta\right)=\beta\left(\mu_{0}+\Delta\right)\), so that \(\beta\left(\mu^{\prime}\right)\) is symmetric about \(\mu_{0}\).

The article "Uncertainty Estimation in Railway Track LifeCycle Cost" (J. of Rail and Rapid Transit, 2009) presented the following data on time to repair (min) a rail break in the high rail on a curved track of a certain railway line. A normal probability plot of the data shows a reasonably linear pattern, so it is plausible that the population distribution of repair time is at least approximately normal. The sample mean and standard deviation are \(249.7\) and \(145.1\), respectively. a. Is there compelling evidence for concluding that true average repair time exceeds \(200 \mathrm{~min}\) ? Carry out a test of hypotheses using a significance level of \(.05\). b. Using \(\sigma=150\), what is the type II error probability of the test used in (a) when true average repair time is actually \(300 \mathrm{~min}\) ? That is, what is \(\beta(300)\) ?

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 \(\mu=95\) ?

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