i=1 X i/n i=1 (X i X) 2 /(n 1). Find the constant c so that the statistic c(x X n+1 )/S has a t-distribution. If n = 8, determine k such that

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Laureen Rodgers
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1 Math 47 Homework Assignment 4 Problem 411 Let X 1, X,, X n, X n+1 be a random sample of size n + 1, n > 1, from a distribution that is N(µ, σ ) Let X = n i=1 X i/n and S = n i=1 (X i X) /(n 1) Find the constant c so that the statistic c(x X n+1 )/S has a t-distribution If n = 8, determine k such that P (X ks < X 9 < X + ks) = 080 The observed interval ( x ks, x + ks) is often called an 80% prediction interval for X 9 Solution 411 The random variable X X n+1 has a normal distribution since it is a linear combination of independent normally distributed random variables Its expected value and variance are E[X X n+1 ] = µ µ = 0, Var[X X n+1 ] = σ n + σ = σ ( n + 1 Therefore, n/(n + 1)(X X n+1 )/σ has a standard normal distribution By Student s theorem, X and S are independent random variables, and (n 1)S /σ has a χ-square distribution with r = n 1 degrees of freedom Since X and S are functions of X 1,, X n, we see that X, S, and X n+1 are independent random variables, therefore X X n+1 and S are independent random variables From these observations we conclude that n/(n + 1)(X Xn+1 )/σ n/(n + 1)(X Xn+1 ) = S /σ S has a t-distribution with r = n 1 degrees of freedom It follows that n c = n + 1 Let n = 8 Then c = 8/9 = /3 Let T 7 be a t-distributed random variable with 7 degrees of freedom and let t 010,7 be the number defined by P (T 7 > t 010,7 ) = 010 Then P ( t 010,7 < T 7 < t 010,7 ) = 080 This, combined with the observation that t 010,7 < c(x X 9) S shows that k = t 010,7 c < t 010,7 X t 010,7 c = 3t 010,7 = (3)(1415) 88 1 n ) S < X 9 < X + t 010,7 S, c = 1501

2 Problem 418 Let X 1, X,, X n be a random sample from N(µ, σ ), where both parameters µ and σ are unknown A confidence interval for σ can be found as follows We know that (n 1)S /σ is a random variable with a χ (n 1) distribution Thus we can find constants a and b so that P ((n 1)S /σ < b) = 0975 and P (a < (n 1)S /σ < b) = 095 (a) Show that this second probability statement can be written as ( (n 1)S P < σ < b ) (n 1) = 095 a (b) If n = 9 and s = 793, find a 95% confidence interval for σ (c) If µ is known, how would you modify the preceding procedure for finding a confidence interval for σ? Solution 418 (a) With the numbers a and b chosen as above, the result follows immediately from the observation that a < (n 1) σ < b (n 1) b < σ < (n 1) a (b) Let W be a χ -distributed random variable with 8 degrees of freedom Then P (W < ) = 0975 and P (1797 < W < ) = 095 With n = 9 we calculate the lower confidence limit to be (n 1)s /b = (8)(793)/(175345) = 3618 and the upper confidence limit to be (n 1)s /a = (8)(793)/(1797) = 9104 Therefore the 95% confidence interval for σ is (3618, 9104) (c) If µ is known then we can replace (n 1)S /σ with n i=1 (X i µ) /σ, which has a χ-square distribution with n degrees of freedom, as a pivotal random variable Problem 45 To illustrate Exercise 44, let X 1,, X 9 and Y 1,, Y 1 represent two independent random samples from the respective normal distributions N(µ 1, σ 1 ) and N(µ, σ ) It is given that σ 1 = 3σ, but σ is unknown Define a random variable that has a t-distribution that can be used to find a 95% confidence interval for µ 1 µ Solution 45 Let X = 9 i=1 X i/9 and Y = 1 i=1 Y i/1 Since X Y is a linear combination of independent normal random variables, it has a

3 3 normal distribution with Thus the random variable (1) E[X Y ] = µ 1 µ Var[X Y ] = σ σ 1 = σ σ 1 ( 1 1 = σ ) 4 (X Y ) (µ 1 µ ) σ 1 1/9 + 1/4 has a standard normal distribution Since 8S1 /σ 1 and 11 /σ = 11 /(3σ 1 ) are independent random variables that have χ -distributions with respective degrees of freedom 8 and 11, the random variable 19S p/σ 1 = 8S 1/σ S /(3σ 1) = (8S S /3)/σ 1 has a χ -distribution with 19 degrees of freedom Therefore S p /σ 1 is the square root of a χ -distributed random variable divided by its degrees of freedom Since Student s Theorem implies that X, Y, S1, and are independent random variables, we see that the random variable in equation (1) and Sp are independent Therefore (X Y ) (µ 1 µ ) σ 1 1/9+1/4 S p /σ 1 = (X Y ) (µ 1 µ ) S p 1/9 + 1/4 has a t-distribution with 19 degrees of freedom that can be used for a 95% confidence interval for µ 1 µ, where S p = (8S /3)/19 Problem 47 Let X 1, X,, X n and Y 1, Y,, Y m be two independent random samples from the respective normal distributions N(µ 1, σ1 ) and N(µ, σ ), where the four parameters are unknown To construct a confidence interval for the ratio, σ1 /σ, of the variances, form the quotient of the two independent χ variables, each divided by its degrees of freedom, namely, () F = (m 1) /(m 1) σ (n 1) /(n 1) σ1 = /σ S1, /σ 1 where S1 and are the respective sample variances (a) What kind of distribution does F have? (b) From the appropriate table, a and b can be found so that P (F < b) = 0975 and P (a < F < b) = 095

4 4 (c) Rewrite the second probability statement as [ ] (3) P a 1 < σ 1 σ < b 1 = 095 The observed values, s 1 and s, can be inserted in these inequalities to provide a 95% confidence interval for σ1 /σ Solution 47 (a) The random variable F in equation () has an F -distribution with r 1 = m numerator degrees of freedom and r = n denominator degrees of freedom (b) Using the notation of Table V on page 661, b = F 005 (m, n) and a = F 0975 (m, n) (c) A direct calculation shows that a < /σ S 1 /σ 1 < b a 1 S < σ 1 σ < b 1 Combining this calculation with P (a < F < b) = 095 leads immediately to equation (3) We therefore find a 95% confidence interval for the ratio σ1 /σ is given by ( F 0975 (m, n) 1, F 0075 (m, n) 1 Problem 465 Assume that the weight of cereal in a 10-ounce box is N(µ, σ) To test H 0 : µ = 101 against H 1 : µ > 101, we take a random sample of size n = 16 and observe that x = 104 and s = 04 ) (a) Do we accept or reject H 0 at the 5% significance level? (b) What is the approximate p-value of this test? Solution 465 (a) The test statistic, 16(X 101)/S has a Student s t-distribution with r = 15 degrees of freedom The realization of the test statistic is t = (4)( )/(04) = 3, and the critical value for the right-tailed alternative H 1 is t 005,15 = 1753 Since t > t005,15, we reject H 0 at the 5% significance level (b) The approximate p-value of this test is p = P (T 15 > 3) = Problem 466 Each of 51 golfers hit three golf balls of brand X and three golf balls of brand Y in a random order Let X i and Y i equal the averages of the distances traveled by the brand X and brand Y golf balls hit by the ith golfer, i = 1,,, 51 Let W i = X i Y i, i = 1,,, 51 Test H 0 : µ W = 0 against H 1 : µ W > 0, where µ W is the mean of the differences If w = 07 and s W = 8463, would H 0 be accepted or rejected at an α = 005 significance level? What is the p-value of this test?

5 Solution 466 We may assume that the distribution of the test statistic (W µ W )/(S W / 51) is approximately the standard normal distribution Under the null hypothesis the realization of the test statistic is z = w 0 s w /n = /51 = 1607 The critical value for the right-tailed alternative hypothesis is z 005 = 1645, and since z < z 005 we accept H 0 at the α = 005 significance level The p-value of this test is p = P (Z > 1607) = 0054 Problem 467 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 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 unknown variances are equal Let α = 005 (b) If x = 79, s x = 56, ȳ = 817, and s y = 83, calculate the value of the test statistic and state your conclusion Solution 467 (a) We are testing a difference in means and the sample sizes are small This suggests using the random variable T 7 = (X Y ) (µ X µ Y ) S p 1/13 + 1/16, where Sp = 1 X + 15 Y, 7 as our test statistic The random variable T 7 has an approximate Student s t-distribution with 7 degrees of freedom We are testing the left-tailed alternative µ X µ Y < 0, which leads to the critical region t < t 005,7, where t 005,7 = 1703 (b) From the given data, the realization of S p is s p = 1s x + 15s y 7 (1)(56) = + (15)(83) 7 and, assuming H 0 is true, the realization of T 7 is t = ( x ȳ) 0 s p 1/13 + 1/16 = (7133)(03734) = 7133 = Since t > 1703, t lies outside the critical region Therefore these data provide insufficient evidence to reject H 0 at the significance level α = 005 The p-value of this test is p = P (T 7 < 1703) =

6 6 Problem 468 Let p equal the proportion of drivers who use a seat belt in a country that does not have a mandatory seat belt law It was claimed that p = 014 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? Solution 468 We will perform a large sample hypothesis test to test whether the value of p increased after the advertising campaign We test whether the data support rejection of the null hypothesis H 0 : p = 014 in favor of the right-tailed alternative hypothesis H a : p > 014 We will use the random variable ˆp p Z =, p(1 p)/n whose distribution is approximately the standard normal distribution, as the test statistic We will test at the significance level of α = 005 The approximate rejection region is z > z 005 = 1645 From the given data, the realized value of ˆp is ˆp = 104/590 = 0176 and, assume the null hypothesis is true, the realized value of Z is z = (014)(1 014)/590 = 539 and the approximate p-value of this test is p = P (Z > 539) = Since z > 1645 these data support rejection of the null hypothesis at the significance level of α = 005 In fact, the p-value shows that these data would support rejection of H 0 in favor of H a even at the significance level of α = 001 These data certainly support the claim that the proportion of seat belt use increased following the advertising campaign Problem 469 In Exercise 418 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 unknown In testing H 0 : σ = σ0 against H 1 : σ > σ0, use the critical region defined by (n 1) /σ0 c That is, reject H 0 and accept H 1 if S cσ0 /(n 1) If n = 13 and the significance level α = 005, determine c Solution 469 Let the random variable W have a χ -distribution with r = n 1 = 1 degrees of freedom By Student s Theorem, 1S /σ0 has the same distribution, therefore c is determined by the property P (W c) = 005, which yields c = 3337 Problem 4610 In Exercise 47, in finding a confidence interval for the ratio of the variances of two normal distributions, we used a statistic S1 /, 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 = 005, find c

7 Solution 4610 By Student s Theorem, the numerator degrees of freedom is r 1 = n 1 = 1 and the denominator degrees of freedom is r = m 1 = 10 The value of c is determined by the property P (F c) = 005 Using the notation of Table V on page 66 we find that c = F 005 (1, 10), and using the program R we find c = 913 7

Solution: First note that the power function of the test is given as follows,

Solution: First note that the power function of the test is given as follows, Problem 4.5.8: Assume the life of a tire given by X is distributed N(θ, 5000 ) Past experience indicates that θ = 30000. The manufacturere claims the tires made by a new process have mean θ > 30000. Is