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Chapter 10: Problem 5

Let \(Y_{1}\) and \(Y_{2}\) be independent and identically distributed with a uniform distribution over the interval \((\theta, \theta+1)\). For testing \(H_{0}: \theta=0\) versus \(H_{a}: \theta>0,\) we have two competing tests: Test 1: Reject \(H_{0}\) if \(Y_{1}>.95\). Test 2: Reject \(H_{0}\) if \(Y_{1}+Y_{2}>c\). Find the value of \(c\) so that test 2 has the same value for \(\alpha\) as test \(1 .\) [Hint: In Example 6.3 , we derived the density and distribution function of the sum of two independent random variables that are uniformly distributed on the interval \((0,1) .]\)

Short Answer

Expert verified
The critical value \(c\) for Test 2 is 1.9.

Step by step solution

01

Define the distribution and problem parameters

The random variables \(Y_1\) and \(Y_2\) are uniformly distributed over the interval \((\theta, \theta + 1)\). The hypothesis test involves testing \(H_0: \theta = 0\) against \(H_a: \theta > 0\). We begin by considering that under \(H_0\), both \(Y_1\) and \(Y_2\) are uniformly distributed over the interval \((0, 1)\).
02

Determine significance level for Test 1

Test 1 rejects \(H_0\) if \(Y_1 > 0.95\). Since \(Y_1\) is uniformly distributed over \((0, 1)\), the probability \(P(Y_1 > 0.95) = 1 - 0.95 = 0.05\). Hence, the significance level \(\alpha\) for Test 1 is 0.05.
03

Find distribution of the sum under null hypothesis

Under \(H_0\), \(Y_1\) and \(Y_2\) are uniformly(0,1), so we need the cumulative distribution function of \(Y_1 + Y_2\). For \(u < 1\), \(P(Y_1 + Y_2 \leq u) = \frac{u^2}{2}\). For \(1 \leq u \leq 2\), \(P(Y_1 + Y_2 \leq u) = u - \frac{u^2}{2} + 0.5\).
04

Calculate critical value \(c\) for Test 2 with the same \(\alpha\)

Since \(\alpha = 0.05\), we find \(c\) such that \(P(Y_1 + Y_2 > c) = 0.05\). Therefore, \(P(Y_1 + Y_2 \leq c) = 0.95\). For \(c > 1\), solve \(c - \frac{c^2}{2} + 0.5 = 0.95\).
05

Solve the equation to find \(c\)

Using the equation, \(c - \frac{c^2}{2} + 0.5 = 0.95\) simplifies to \(c^2 - 2c + 0.1 = 0\). Solving this quadratic equation, we use the quadratic formula \(c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -2\), \(c = 0.1\). Substituting values, we get \(c = \frac{2 \pm \sqrt{3.6}}{2}\). The positive value \(c = 1.9\).

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Key Concepts

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

Uniform Distribution
Uniform Distribution is a fundamental concept in probability and statistics. Imagine distributing a set of values over an interval, such as between 0 and 1, so that each value has an equal probability of occurring. This is what we call a uniform distribution. Specifically, when a random variable is "uniformly distributed," it means that any segment of a given interval is equally likely to contain the value of the variable.
  • Each outcome is equally likely, meaning the probability is constant across the interval.
  • The density function is flat, indicating that there's no preference for any particular value within the interval.
  • In the problem, the variables "Y1" and "Y2" are uniformly distributed over the interval \((\theta, \theta + 1)\). Under the null hypothesis \(H_0\), their distribution is defined over \((0, 1)\).
Understanding uniform distribution is critical for setting up and solving hypothesis testing related to such variables.
Significance Level
The "Significance Level," often denoted by \(\alpha\), is a key concept in hypothesis testing. It represents the probability of rejecting the null hypothesis when it is actually true. This is commonly known as making a Type I error. The significance level is chosen before conducting the test and reflects the level of risk you're willing to accept for drawing a wrong conclusion.
  • In hypothesis testing, a lower significance level means you're more cautious about rejecting the null hypothesis.
  • Typical significance levels are 0.05, 0.01, or even 0.10 depending on the context and field of study.
  • In the exercise, Test 1's significance level is set at \(0.05\) because \(Y_1 > 0.95\) leads to a probability of 0.05, implying a 5% chance of making an error by rejecting \(H_0\).
By aligning Test 2 with the same significance level, you ensure both tests are equally stringent in their criteria for rejecting \(H_0\).
Cumulative Distribution Function
The Cumulative Distribution Function (CDF) gives us vital information about the probability that a random variable will take a value less than or equal to a specific number. In other words, the CDF encapsulates the spread of the distribution across different values. For uniformly distributed variables, the CDF functions differ based on where the sum \(Y_1 + Y_2\) falls within particular ranges.
  • If the sum is less than 1, the probability is expressed by \(\frac{u^2}{2}\).
  • If the sum is between 1 and 2, the probability is articulated as \(u - \frac{u^2}{2} + 0.5\).
  • These CDF segments are crucial for determining the critical value \(c\) in Test 2.
By solving \(c - \frac{c^2}{2} + 0.5 = 0.95\) through the CDF, we approximate \(c\) such that the probability \(Y_1 + Y_2 > c\) equals the significance level of Test 1, ensuring the tests' comparability.

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Most popular questions from this chapter

Researchers have shown that cigarette smoking has a deleterious effect on lung function. In their study of the effect of cigarette smoking on the carbon monoxide diffusing capacity (DL) of the lung, Ronald Knudson, W. Kaltenborn and B. Burrows found that current smokers had DL readings significantly lower than either ex-smokers or nonsmokers. \(^{*}\) The carbon monoxide diffusing capacity for a random sample of current smokers was as follows: $$\begin{array}{rrrrr} 103.768 & 88.602 & 73.003 & 123.086 & 91.052 \\ 92.295 & 61.675 & 90.677 & 84.023 & 76.014 \\ 100.615 & 88.017 & 71.210 & 82.115 & 89.222 \\ 102.754 & 108.579 & 73.154 & 106.755 & 90.479 \end{array}$$ Do these data indicate that the mean DL reading for current smokers is lower than 100 , the average DL reading for nonsmokers? a. Test at the \(\alpha=.01\) level. b. Bound the \(p\) -value using a table in the appendix. c. Find the exact \(p\) -value.

From two normal populations with respective variances \(\sigma_{1}^{2}\) and \(\sigma_{2}^{2},\) we observe independent sample variances \(S_{1}^{2}\) and \(S_{2}^{2}\), with corresponding degrees of freedom \(\nu_{1}=n_{1}-1\) and \(\nu_{2}=n_{2}-1 .\) We wish to test \(H_{0}: \sigma_{1}^{2}=\sigma_{2}^{2}\) versus \(H_{a}: \sigma_{1}^{2} \neq \sigma_{2}^{2}\) a. Show that the rejection region given by $$\left\\{F>F_{\nu_{2}, \alpha / 2}^{\nu_{1}} \quad \text { or } \quad F<\left(F_{\nu_{1}, \alpha / 2}^{\nu_{2}}\right)^{-1}\right\\}$$ where \(F=S_{1}^{2} / S_{2}^{2},\) is the same as the rejection region given by $$\left\\{S_{1}^{2} / S_{2}^{2}>F_{\nu_{2}, \alpha / 2}^{\nu_{2}} \text { or } S_{2}^{2} / S_{1}^{2}>F_{\nu_{1}, \alpha / 2}^{\nu_{2}}\right\\}$$ b. Let \(S_{L}^{2}\) denote the larger of \(S_{1}^{2}\) and \(S_{2}^{2}\) and let \(S_{S}^{2}\) denote the smaller of \(S_{1}^{2}\) and \(S_{2}^{2} .\) Let \(\nu_{L}\) and \(\nu_{S}\) denote the degrees of freedom associated with \(S_{L}^{2}\) and \(S_{S}^{2}\), respectively. Use part (a) to show that, under \(H_{0}\) $$\mathrm{P}\left(S_{L}^{2} / S_{S}^{2}>F_{\nu_{S}, \alpha / 2}^{\mu_{L}}\right)=\alpha$$ Notice that this gives an equivalent method for testing the equality of two variances.

The stability of measurements of the characteristics of a manufactured product is important in maintaining product quality. In fact, it is sometimes better to obtain small variation in the measured value of some important characteristic of a product and have the process mean slightly off target than to get wide variation with a mean value that perfectly fits requirements. The latter situation may produce a higher percentage of defective product than the former. A manufacturer of light bulbs suspected that one of his production lines was producing bulbs with a high variation in length of life. To test this theory, he compared the lengths of life of \(n=50\) bulbs randomly sampled from the suspect line and \(n=50\) from a line that seemed to be in control. The sample means and variances for the two samples were as shown in the following table. $$\begin{array}{ll} \hline \text { Suspect Line } & \text { Line in Control } \\ \hline \bar{y}_{1}=1,520 & \bar{y}_{2}=1,476 \\ s_{1}^{2}=92,000 & s_{2}^{2}=37,000 \\ \hline \end{array}$$ a. Do the data provide sufficient evidence to indicate that bulbs produced by the suspect line possess a larger variance in length of life than those produced by the line that is assumed to be in control? Use \(\alpha=.05\) b. Find the approximate observed significance level for the test and interpret its value.

Let \(X_{1}, X_{2}, \ldots, X_{m}\) denote a random sample from the exponential density with mean \(\theta_{1}\) and let \(Y_{1}, Y_{2}, \ldots, Y_{n}\) denote an independent random sample from an exponential density with \(\theta_{2}\). a. Find the likelihood ratio criterion for testing \(H_{0}: \theta_{1}=\theta_{2}\) versus \(H_{a}: \theta_{1} \neq \theta_{2}\). b. Show that the test in part (a) is equivalent to an exact \(F\) test [Hint: Transform \(\sum X_{i}\) and \(\sum Y_{j}\) to \(\left.\chi^{2} \text { random variables. }\right]\)

An experimenter has prepared a drug dosage level that she claims will induce sleep for \(80 \%\) of people suffering from insomnia. After examining the dosage, we feel that her claims regarding the effectiveness of the dosage are inflated. In an attempt to disprove her claim, we administer her prescribed dosage to 20 insomniacs and we observe \(Y\), the number for whom the drug dose induces sleep. We wish to test the hypothesis \(H_{0}: p=.8\) versus the alternative, \(H_{a}: p<.8 .\) Assume that the rejection region \(\\{y \leq 12\\}\) is used. a. In terms of this problem, what is a type I error? b. Find \(\alpha\) c. In terms of this problem, what is a type II error? d. Find \(\beta\) when \(p=.6\) e. Find \(\beta\) when \(p=.4\)
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