MATH 78 Homewor 3 Olesandr Pavleno 4.5.8 Let us say the life of a tire in miles, say X, is normally distributed with mean θ and standard deviation 5000. Past experience indicates that θ = 30000. The manufacturer claims that the tires made by a new process have mean θ > 30000. It is possible that θ = 35000. Chec his claim by testing H 0 : θ = 30000 against H 1 : θ > 30000. We shall observe n independent values of X, say x 1,..., x n, and we shall reject H 0 (thus accept H 1 if and only if x c. Determine n and c so that the power function γ(θ of the test has the values γ(30000 = 0.01 and γ(35000 = 0.98. Solution. To determine n and c we use given conditions for the power function and critical values of normal distribution: c 30000 5000/ n = z 0.01 =.36 and c 35000 5000/ n = z 0.0 =.054. Solving the system we obtain c = 3655.3 and n = 19.184 19. Then, we reject H 0 (thus accept H 1 if and only if x 3655.3. 4.5.9 Let X have a Poisson distribution with mean θ. Consider the simple hypothesis H 0 : θ = 1/ and the alternative composite hypothesis H 1 : θ < 1/. Thus Ω = {θ : 0 < θ 1/}. Let X 1,..., X 1 denote a random sample of size 1 from this distribution. We reject H 0 if and only if the observed value of Y = X 1 +... + X 1. If γ(θ is a power function of the test, find the powers γ(1/, γ(1/3, γ(1/4, γ(1/6, and γ(1/1. Setch the graph of γ(θ. What is the significance level of the test? Solution. If random variables X 1,..., X 1 Poisson(θ, then Y = X 1 +... + X 1 Poisson(1 θ. Hence, for θ = 1/ Y Poisson(6 and γ(1/ = P (Y = For θ = 1/3 Y Poisson(4 and γ(1/3 = P (Y = For θ = 1/4 Y Poisson(3 and γ(1/4 = P (Y = 6! e 6 = e 6 4! e 4 = e 4 3! e 3 = e 3 1 (1 + 6 + 6 = 0.0619. (1 + 4 + 4 = 0.38. (1 + 3 + 3 = 0.43.
For θ = 1/6 Y Poisson( and γ(1/6 = P (Y =! e = e (1 + + = 0.677. For θ = 1/1 Y Poisson(1 and γ(1/1 = P (Y = ( 1! e 1 = e 1 1 + 1 + 1 = 0.919. Significance level of the test is γ(1/ = 0.0619 = 6.%. 4.5.10 Let Y have a binomial distribution with parameters n and p. We reject H 0 : p = 1/ and accept H 1 : p > 1/ if Y c. Find n and c to give a power function γ(p which is such that γ(1/ = 0.1 and γ(/3 = 0.95, approximately. Solution. To determine n and c we use given conditions for the power function and critical values of normal distribution: c np = c n/ = z 0.1 = 1.8 np(1 p n/ and c n/3 n/3 = z 0.05 = 1.645. Solving the system we obtain c = 41.56 and n = 7.5 7. Then power function of the test is ( Y np γ(p = P (Y c = P c np = np(1 p np(1 p = P ( Y 7p 7p(1 p 41.56 7p, 1/ p < 1. 7p(1 p
4.5.13 Let p denote the probability that, for a particular tennis player, the first serve is good. Since p = 0.4, this player decided to tae lessons in order to increase p. When the lessons are completed, the hypothesis H 0 : p = 0.4 will be tested against H 1 : p > 0.4 based on n = 5 trials. Let y equal the number of first serves that are good, and let the critical region be defined by C = {y : y 13}. (a Determine α = P (Y 13; p = 0.4. Solution. α = P (Y 13; p = 0.4 = = 5 =13 n =13 ( n p (1 p n = ( 5 (0.4 (0.6 5 = 0.1537. (b Find β = P (Y < 13 when p = 0.6 ; that is, β = P (Y 1; p = 0.6 so that 1 β is the power at p = 0.6. 1 ( n β = P (Y 1; p = 0.6 = p (1 p n = =1 =1 1 ( 5 = (0.6 (0.4 5 = 0.1537. 4.6.4 Consider the one-sided t -test for H 0 : µ = µ 0 versus H A1 : µ > µ 0 constructed in Ex. 5.5.4 and the two-sided t -test for H 0 : µ = µ 0 versus H 0 : µ µ 0 given in (5.6.9. Assume that both tests are of size α. Show that for µ > µ 0, the power function of the one-sided test is larger than the power function of the two-sided test. Solution. For both tests X n N(µ, σ. The first test s hypotheses are H 0 : µ = µ 0 µ µ and H 1 : µ > µ 0, and we reject H 0 if 0 σ/ t n α,n 1. The second test s hypotheses are H 0 : µ = µ 0 and H 1 : µ µ 0, and we reject H 0 if µ µ 0 σ/ n t α/,n 1. Then, since t α,n 1 < t α/,n 1, for µ > µ 0 ( ( µ µ 0 γ (µ = P σ/ µ n t µ0 α/,n 1 = P σ/ n t α/,n 1 ( µ µ0 P σ/ n t α,n 1 = γ 1 (µ. 3
4.6.5 Assume that the weight of cereal in a "10-ounce box"is N(µ, σ. To test H 0 : µ = 10.1 against H 1 : µ > 10.1, we tae a random sample of size n = 16 and observe that x = 10.4 and s = 0.4. (a Do we accept or reject H 0 at the 5% significance level? Solution. t obs = x µ 0 s/ n = 10.4 10.1 0.4/4 Then, since t obs = 3 > t 0.05,15 = 1.753, we reject H 0 and accept H 1 at the 5% significance level. (b What is the approximate p -value of this test? To determine p -value of the test we have to find α such that t α,15 3. Using table with critical values of t distribution, p -value = α = 0.005. = 3. 4.6.7 Among the data collected for the World Health Organization air quality monitoring project is a measure of suspended particles in µg/m 3. Let X and Y equal the concentration of suspended particles in µg/m 3 in the city center (commercial district for Melbourne and Houston, respectively. Using n = 13 observations of X and m = 16 observations of Y, we shall test H 0 : µ X = µ Y against H 1 : µ X < µ Y. (a Define the test statistic and critical region, assuming that the unnown variances are equal. Let α = 0.05. and Solution. The test statistic and the critical region will be Y X (n T = 1 1S1 + (n 1S, where S p =, S p 1/n1 + 1/n n 1 + n C = {(X 1,..., X 13, Y 1,..., Y 16 : T t 0.05,7 = 1.703}. (b If x = 7.9, s x = 5.6, y = 81.7, and s y = 8.3, calculate the value of the test statistic and state your conclusion. and S p = T obs = 1(5.6 + 15(8.3 7 = 7.133, 81.7 7.9 7.133 1/13 + 1/16 = 0.8685. 4
Then, since T abs < t 0.05,7 = 1.703, we do not reject H 0 at the significance level α = 0.05. 4.6.8 Let p equal the proportion of drivers who use a seat belt in a state that does not have a mandatory seat belt law. It was claimed that p = 0.14. An advertising campaign was conducted to increase this proportion. Two months after the campaign, y = 104 out of a random sample of n = 590 drivers were wearing their seat belts. Was the campaign successful? (a Define the null and alternative hypotheses. Solution. Hypotheses are H 0 : p = 0.14 against H 1 : p > 0.14. (b Define a critical region with an α = 0.01 significance level. The critical region is { C = y : p p 0 p0 (1 p 0 /n z 0.01 =.36 }, where p = y n. (c Determine the approximate p -value and state your conclusion. z obs = p p 0 p0 (1 p 0 /n = 0.1763 0.14 0.14(0.86/590 =.539. Then, using the table with critical values of normal distribution, correspondent p - value = 0.0055. And, since 0.0055 < α = 0.01, we have to reject H 0 at the significance level α = 0.01. 4.6.9 In exercise 5.4.14 we found a confidence interval for the variance σ using the variance S of a random sample of size n arising from N(µ, σ, where the mean µ is unnown. In testing H 0 : σ = σ0 against H 1 : σ > σ0, use the critical region defined by (n 1S /σ0 c. That is, reject H 0 and accept H 1 if S cσ0(n 1. If n = 13 and the significance level α = 0.05, determine c. Solution. Since (n 1S σ 0 χ (n 1, given mentioned conditions, c = χ 0.05(1 = 4.404. 5
4.6.10 In exercise 5.4.5, in finding a confidence interval for the ratio of the variances of two normal distributions, we used a statistic S1/S which has an F-distribution when those two variances are equal. If we denote that statistic by F, we can test H 0 : σ1 = σ against H 1 : σ1 > σ using the critical region F c. If n = 13, m = 11, and α = 0.05, find c. Solution. Since S1/S F α (n 1, m 1, given mentioned conditions, c = F (1, 10; 0.05 =.91. 6