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Chapter 9: Problem 30

Suppose a level \(.05\) test of \(H_{0}: \mu_{1}-\mu_{2}=0\) versus \(H_{a}: \mu_{1}-\mu_{2}>0\) 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
Calculating \( \beta \), probabilities vary with \( n \), larger samples yield smaller \( \beta \). For small effect size, large \( n \) may not be practical.

Step by step solution

01

Define Type I and Type II Errors

A Type I error occurs when we reject the null hypothesis when it is actually true. The significance level \( \alpha = 0.05 \) specifies the probability of making a Type I error in this test. A Type II error occurs when we fail to reject the null hypothesis when the alternative hypothesis is true. The probability of a Type II error is denoted by \( \beta \). Our task is to calculate this for different sample sizes \( n \).
02

Calculate Critical Value

Given the significance level \( \alpha = 0.05 \) for a one-tailed test, the critical value corresponds to the \( 95\% \) percentile of the normal distribution, as the test is right-tailed. Therefore, the critical value \( z_{\alpha} \) is approximately 1.645.
03

Compute Test Statistic Under \( H_{a} \)

The test statistic for the given means under the alternative hypothesis \( H_{a}: \mu_{1} - \mu_{2} = 1 \) is calculated using:\[ z = \frac{(\mu_{1} - \mu_{2}) - 0}{\sigma \sqrt{\frac{2}{n}}} \]Substituting the known values of \( \sigma_{1} = \sigma_{2} = 10 \), we have \( \sigma = 10 \):\[ z = \frac{1}{10 \sqrt{\frac{2}{n}}} \]
04

Calculate Probability of Type II Error

For Type II error (\( \beta \)), we find the probability that the test statistic is less than the critical value \( z_{\alpha}=1.645 \) under \( H_{a} \). Calculate \( \beta \) using:For each \( n \):- \( n=25 \): Substitute in earlier test statistic formula:\[ z = \frac{1}{10 \sqrt{\frac{2}{25}}} = \frac{1}{4} \approx 0.25 \]\( \beta = P(Z < 1.645 - 0.25) = P(Z < 1.395) \)- \( n=100 \): \[ z = \frac{1}{10 \sqrt{\frac{2}{100}}} = \frac{1}{2} \approx 0.5 \]\( \beta = P(Z < 1.645 - 0.5) = P(Z < 1.145) \)- \( n=2500 \): \[ z = \frac{1}{10 \sqrt{\frac{2}{2500}}} = \frac{1}{10} = 0.1 \]\( \beta = P(Z < 1.645 - 0.1) = P(Z < 1.545) \)- \( n=10000 \): \[ z = \frac{1}{10 \sqrt{\frac{2}{10000}}} = \frac{1}{20} = 0.05 \]\( \beta = P(Z < 1.645 - 0.05) = P(Z < 1.595) \) Use standard normal distribution tables or tools to find these probabilities.
05

Interpret the Practical Significance

If a difference \( \mu_{1} - \mu_{2} = 1 \) has little practical significance in a real-world scenario, then having a large sample size, such as \( n = 10000 \), might not be desirable as it only distinguishes minute differences that may not bear practical importance. Considerations of cost and practicality often weigh against using very large samples unless the context justifies very accurate detection of small effects.

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Key Concepts

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

Hypothesis Testing
Hypothesis testing is a critical method in statistics that allows us to make decisions or inferences about population parameters based on sample data. It involves two primary hypotheses:
  • Null Hypothesis (\( H_0 \)): This is a statement of no effect or no difference. For example, in the exercise, the null hypothesis is \( \mu_1 - \mu_2 = 0 \), indicating no difference between the two population means.
  • Alternative Hypothesis (\( H_a \)): This suggests there is an effect or a difference. In this case, \( \mu_1 - \mu_2 > 0 \), suggesting one mean is greater than the other.
Hypothesis tests can lead to two types of errors:
  • Type I Error: Rejecting the null hypothesis when it is actually true.
  • Type II Error: Failing to reject the null hypothesis when the alternative is true.
The choice between these affects the test鈥檚 reliability. Understanding and choosing appropriate hypotheses is essential for effective decision-making in research and data analysis.
Significance Level
The significance level (\( \alpha \)) is a crucial concept in hypothesis testing. It represents the probability of committing a Type I error. In simple terms, it's the risk we are willing to take in mistakenly rejecting the null hypothesis.
For the exercise, the significance level is set at \( 0.05 \), which implies:
  • There is a 5% chance of incorrectly concluding there is a difference between the population means when there is not.
This threshold dictates the position of the critical value in a normal distribution:
  • A one-tailed test at \( \alpha = 0.05 \) corresponds to a critical z-value of approximately 1.645, representing the 95th percentile.
Choosing an \( \alpha \) should balance the consequences of both Type I and Type II errors, and often depends on the context, such as the potential impact of an incorrect decision.
Sample Size Calculation
Determining the appropriate sample size is crucial in minimizing errors and enhancing the reliability of hypothesis testing results. Larger sample sizes generally provide more precise estimates and decrease the probability of a Type II error, denoted by \( \beta \).
This exercise demonstrates that as the sample size (\( n \)) increases:
  • The test statistic becomes larger, moving away from the null hypothesis expectation.
  • The probability of \( \beta \) decreases, reflecting increased power (the likelihood of correctly rejecting \( H_0 \)).
For example:
  • At \( n = 25 \), the difference in means of 1 presents substantial practical significance, necessitating careful examination of sample size impacts.
  • Conversely, \( n = 10000 \) might offer little benefit if the difference of 1 is trivial, as it can lead to unnecessary complexity and resource use.
Thus, calculating the right sample size involves weighing statistical power, practical significance, and resources, ensuring the test is efficient and meaningful.

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Most popular questions from this chapter

It has been estimated that between 1945 and 1971 , as many as 2 million children were born to mothers treated with diethylstilbestrol \((\mathrm{DES})\), a nonsteroidal estrogen recommended for pregnancy maintenance. The FDA banned this drug in 1971 because research indicated a link with the incidence of cervical cancer. The article "Effects of Prenatal Exposure to Diethylstilbestrol (DES) on Hemispheric Laterality and Spatial Ability in Human Males" (Hormones and Behavior, 1992: 62-75) discussed a study in which 10 males exposed to \(\mathrm{DES}\), and their unexposed brothers, underwent various tests. This is the summary data on the results of a spatial ability test: \(\bar{x}=12.6\) (exposed), \(\bar{y}=13.7\), and standard error of mean difference \(=.5\). Test at level .05 to see whether exposure is associated with reduced spatial ability by obtaining the P-value.

Cushing's disease is characterized by muscular weakness due to adrenal or pituitary dysfunction. To provide effective treatment, it is important to detect childhood Cushing's disease as early as possible. Age at onset of symptoms and age at diagnosis (months) for 15 children suffering from the disease were given in the article "Treatment of Cushing's Disease in Childhood and Adolescence by Transphenoidal Microadenomectomy" (New Engl. J. of Med., 1984:889). Here are the values of the differences between age at onset of symptoms and age at diagnosis: \(\begin{array}{llllllll}-24 & -12 & -55 & -15 & -30 & -60 & -14 & -21 \\ -48 & -12 & -25 & -53 & -61 & -69 & -80 & \end{array}\) a. Does the accompanying normal probability plot cast strong doubt on the approximate normality of the population distribution of differences? b. Calculate a lower \(95 \%\) confidence bound for the population mean difference, and interpret the resulting bound. c. Suppose the (age at diagnosis) - (age at onset) differences had been calculated. What would be a \(95 \%\) upper confidence bound for the corresponding population mean difference?

Sometimes experiments involving success or failure responses are run in a paired or before/after manner. Suppose that before a major policy speech by a political candidate, \(n\) individuals are selected and asked whether \((S)\) or not \((F)\) they favor the candidate. Then after the speech the same \(n\) people are asked the same question. The responses can be entered in a table as follows: After \(S \quad F\) where \(x_{1}+x_{2}+x_{3}+x_{4}=n\). Let \(p_{1}, p_{2}, p_{3}\), and \(p_{4}\) denote the four cell probabilities, so that \(p_{1}=P(S\) before and \(S\) after), and so on. We wish to test the hypothesis that the true proportion of supporters \((S)\) after the speech has not increased against the alternative that it has increased. a. State the two hypotheses of interest in terms of \(p_{1}, p_{2}, p_{3}\), and \(p_{4}\). b. Construct an estimator for the after/before difference in success probabilities. c. When \(n\) is large, it can be shown that the rv \(\left(X_{f}-X_{j}\right) / n\) has approximately a normal distribution with variance given by \(\left[p_{i}+p_{j}-\left(p_{j}-p_{j}\right)^{2}\right] / n\). Use this to construct a test statistic with approximately a standard normal distribution when \(H_{0}\) is true (the result is called McNemar's test). d. If \(x_{1}=350, \quad x_{2}=150, \quad x_{3}=200\), and \(x_{4}=300\), what do you conclude?

The authors of the article "Dynamics of Canopy Structure and Light Interception in Pinus elliotti, North Florida" (Ecological Monographs, 1991: 33-51) planned an experiment to determine the effect of fertilizer on a measure of leaf area. A number of plots were available for the study, and half were selected at random to be fertilized. To ensure that the plots to receive the fertilizer and the control plots were similar, before beginning the experiment tree density (the number of trees per hectare) was recorded for eight plots to be fertilized and eight control plots, resulting in the given data. Minitab output follows. \(\begin{array}{lrrrrr}\text { Fertilizer plots } & 1024 & 1216 & 1312 & 1280 \\ & 1216 & 1312 & 992 & 1120 \\ \text { Control plots } & 1104 & 1072 & 1088 & 1328 \\ & 1376 & 1280 & 1120 & 1200 \\ & & & & & \\ \text { Two sample } T \text { for fertilizer vs control } \\ & \mathbb{N} & \text { Mean } & \text { StDev } & \text { SE Mean } \\ \text { fertilize } & 8 & 1184 & 126 & 44 \\ \text { control } & 8 & 1196 & 118 & 42 \\\ 95 \% \text { CI for mu fertilize } & -\text { mu control: } \text { (-144, } \\ 120) & & & & & \end{array}\) a. Construct a comparative boxplot and comment on any interesting features. b. Would you conclude that there is a significant difference in the mean tree density for fertilizer and control plots? Use \(\alpha=05\) c. Interpret the given confidence interval.

The following data refers to airborne bacteria count (number of colonies \(/ \mathrm{ft}^{3}\) ) both for \(m=8\) carpeted hospital rooms and for \(n=8\) uncarpeted rooms ("Microbial Air Sampling in a Carpeted Hospital," J. of Environmental Health, 1968: 405). Does there appear to be a difference in true average bacteria count between carpeted and uncarpeted rooms? \(\begin{array}{lrrrrrrrr}\text { Carpeted } & 11.8 & 8.2 & 7.1 & 13.0 & 10.8 & 10.1 & 14.6 & 14.0 \\ \text { Uncarpeted } & 12.1 & 8.3 & 3.8 & 7.2 & 12.0 & 11.1 & 10.1 & 13.7\end{array}\) Suppose you later learned that all carpeted rooms were in a veterans' hospital, whereas all uncarpeted rooms were in a children's hospital. Would you be able to assess the effect of carpeting? Comment.
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