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Chapter 9: Problem 31

Suppose a level 05 test of \(H_{0}: \mu_{1}-\mu_{2}=0\) versus \(H_{\mathrm{a}}: \mu_{1}-\mu_{2}>0\) is to be performed, assuming \(\sigma_{1}=\sigma_{2}=\) 10 and normality of both distributions, using equal sample sizes \((m=n)\). Evaluate the probability of a type II error when \(\mu_{1}-\mu_{2}=1\) and \(n=25,100,2500\), and 10,000 . Can you think of real problems in which the difference \(\mu_{1}-\mu_{2}=1\) has little practical significance? Would sample sizes of \(n=10,000\) be desirable in such problems?

Short Answer

Expert verified
Type II error probabilities are around 0.90 (n=25), 0.76 (n=100), 0.40 (n=2500), 0.16 (n=10,000). Small differences may not justify large samples.

Step by step solution

01

Define Parameters and Hypotheses

Identify the null hypothesis \(H_0\) and the alternative hypothesis \(H_a\). Here, \(H_0: \mu_1 - \mu_2 = 0\) and \(H_a: \mu_1 - \mu_2 > 0\). Also note the given parameters: \(\sigma_1 = \sigma_2 = 10\), \(\mu_1 - \mu_2 = 1\), and sample sizes \(n = 25, 100, 2500, \text{and } 10,000\).
02

Calculate the Test Statistic

Compute the standard error of the difference between the sample means using \(SE(\bar{X}_1 - \bar{X}_2) = \sigma \sqrt{\frac{2}{n}}\). For \(n = 25, SE = 10 \times \sqrt{\frac{2}{25}} = 2.828\). Repeat for \(n = 100, 2500, \text{and } 10,000\), which gives 1.414, 0.2, and 0.0632 respectively.
03

Determine Critical Value

At a significance level of 0.05 for a one-tailed test, the critical value \(z_c\) from the standard normal distribution is 1.645.
04

Calculate the Type II Error Probability (尾)

The type II error (\(\beta\)) probability occurs when the true mean difference is 1 but the null hypothesis is not rejected. Compute \(\beta = P(Z < z_c - \frac{\mu_1 - \mu_2}{SE})\) for each \(n\). For \(n = 25\), \( \beta = P(Z < 1.645 - \frac{1}{2.828}) = P(Z < 1.292) \approx 0.9015\). Calculate similarly for other \(n\) values: \(n = 100\), \(\beta \approx 0.762\); \(n = 2500\), \(\beta \approx 0.403\); \(n = 10,000\), \(\beta \approx 0.156\).
05

Discuss Practical Significance

Assess whether a difference of \(\mu_1 - \mu_2 = 1\) is of practical importance. Often, detecting very small differences (like 1 unit) in practical problems doesn鈥檛 significantly impact decision-making. Large \(n\) like 10,000 should be weighed against costs and pragmatic value.

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Key Concepts

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

Type II Error
Understanding a Type II error is crucial in hypothesis testing. In simple terms, a Type II Error occurs when the test fails to reject a false null hypothesis. This means that we incorrectly conclude there is no effect or difference when one actually exists.
For instance, in the exercise provided, you're looking at a scenario where we assume \((\mu_1 - \mu_2 = 1)\). If our test does not reject the null hypothesis \(H_0:\ \mu_1 - \mu_2 = 0\), despite the fact that \(\mu_1 - \mu_2\) indeed equals 1, a Type II error happens.
The ability to measure Type II error probability, often denoted by \(\beta\), informs us about the power of our test, as \(Power = 1 - \beta\). An understanding of this concept helps in determining the effectiveness of a test, including aspects like detecting small yet significant differences between groups.
Sample Size
The sample size, denoted as \(n\), plays a crucial role in the effectiveness of a hypothesis test. Larger sample sizes generally provide more reliable estimates and increase the power of a test.
From the original exercise, different sample sizes (25, 100, 2500, and 10,000) were used to evaluate their impact on Type II errors. In simple terms, as the sample size increases:
  • The standard error decreases, indicating more precise estimates of the sample means.
  • The probability of a Type II error (\(\beta\)) decreases, enhancing the power to detect the true effect.
However, it is essential to balance the benefits of a larger sample size against practical considerations such as cost, time, and resource availability. In cases where \(\mu_1 - \mu_2 = 1\) lacks practical significance, unnecessarily large sample sizes may not be judicious.
Significance Level
The significance level, usually denoted by \(\alpha\), is a threshold set for determining when to reject the null hypothesis. It reflects the allocator's tolerance for making a Type I error 鈥� which is rejecting a true null hypothesis.
In the context of the provided exercise, a significance level of 0.05 is utilized. This implies that there is a 5% risk of incorrectly rejecting a true null hypothesis.
  • The choice of \(\alpha\) affects both Type I and Type II errors. Lowering \(\alpha\) reduces the probability of a Type I error but may increase the \(\beta\) (Type II error probability).
  • Decision-makers must prioritize depending on whether they aim to minimize Type I or Type II errors in their specific context.
Thus, selecting an appropriate significance level requires careful consideration of the test's objective and the potential consequences of errors. Trade-offs between Type I and Type II errors are part of designing an efficient hypothesis test.

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Most popular questions from this chapter

Tensile-strength tests were carried out on two different grades of wire rod ('"Fuidized Bed Patenting of Wire Rods, Wire \(J .\) J June 1977: \(56-61\) ), resulting in the accompanying data. \begin{tabular}{lccc} Grade & Sample Size & Sample Mean \(\left(\mathrm{kg} / \mathrm{mm}^{2}\right)\) & Sample SD \\ \hline AISI 1064 & \(m=129\) & \(\bar{x}=107.6\) & \(s_{1}=1.3\) \\ AISI 1078 & \(n=129\) & \(\bar{y}=123.6\) & \(s_{2}=2.0\) \\ \hline \end{tabular} a. Does the data provide compelling evidence for concluding that true average strength for the 1078 grade exceeds that for the 1064 grade by more than \(10 \mathrm{~kg} /\) \(\mathrm{mm}^{2}\) ? Test the appropriate hypotheses using a significance level of \(.01\). b. Estimate the difference between true average strengths for the two grades in a way that provides information about precision and reliability.

The level of lead in the blood was determined for a sample of 152 male hazardous-waste workers ages 20-30 and also for a sample of 86 female workers, resulting in a mean \(\pm\) standard error of \(5.5 \pm 0.3\) for the men and \(3.8 \pm 0.2\) for the women (" Temporal Changes in Blood Lead Levels of Hazardous Waste Workers in New Jersey, 1984-1987," Environ. Monitoring and Assessment, 1993: 99-107). Calculate an estimate of the difference between true average blood lead levels for male and female workers in a way that provides information about reliability and precision.

Determine the number of degrees of freedom for the twosample \(t\) test or CI in each of the following situations: a. \(m=10, n=10, s_{1}=5.0, s_{2}=6.0\) b. \(m=10, n=15, s_{1}=5.0, s_{2}=6.0\) c. \(m=10, n=15, s_{1}=2.0, s_{2}=6.0\) d. \(m=12, n=24, s_{1}=5.0, s_{2}=6.0\)

Fusible interlinings are being used with increasing frequency to support outer fabrics and improve the shape and drape of various pieces of clothing. The article "Compatibility of Outer and Fusible Interlining Fabrics in Tailored Garments" (Textile Res. J., 1997: \(137-142\) ) gave the accompanying data on extensibility (\%) at \(100 \mathrm{gm} / \mathrm{cm}\) for both high-quality (H) fabric and poor-quality (P) fabric specimens. \(\begin{array}{rrrrrrrrrr}\mathrm{H} & 1.2 & .9 & .7 & 1.0 & 1.7 & 1.7 & 1.1 & .9 & 1.7 \\ & 1.9 & 1.3 & 2.1 & 1.6 & 1.8 & 1.4 & 1.3 & 1.9 & 1.6 \\ & .8 & 2.0 & 1.7 & 1.6 & 2.3 & 2.0 & & & \\ \mathrm{P} & 1.6 & 1.5 & 1.1 & 2.1 & 1.5 & 1.3 & 1.0 & 2.6 & \end{array}\) a. Construct normal probability plots to verify the plausibility of both samples having been selected from normal population distributions. b. Construct a comparative boxplot. Does it suggest that there is a difference between true average extensibility for high-quality fabric specimens and that for poor-quality specimens? c. The sample mean and standard deviation for the highquality sample are \(1.508\) and \(.444\), respectively, and those for the poor-quality sample are \(1.588\) and \(.530\). Use the two-sample \(t\) test to decide whether true average extensibility differs for the two types of fabric.

Give as much information as you can about the \(P\)-value of the \(F\) test in each of the following situations: a. \(v_{1}=5, v_{2}=10\), upper-tailed test, \(f=4.75\) b. \(v_{1}=5, v_{2}=10\), upper-tailed test, \(f=2.00\) c. \(v_{1}=5, v_{2}=10\), two-tailed test, \(f=5.64\) d. \(v_{1}=5, v_{2}=10\), lower-tailed test, \(f=.200\) e. \(v_{1}=35, v_{2}=20\), upper-tailed test, \(f=3.24\)
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