Tutorial 8 Solutions

UNIVERSITY OF CAPE TOWN DEPARTMENT OF STATISTICAL SCIENCES STA2030S Tutorial 8 Solutions 2. a) Y is a statistic. It does not depend on a population parameter b) Z is NOT a statistic. It depends on μ and σ c) S 2 is a statistic. It does not depend on a population parameter d) Z is NOT a statistic. It depends on μ 3. We have X i ∼ N ( μ, σ 2 ) and ∑ 16 i =1 ( X i - ¯ X ) 2 = 133. The hypotheses for this two-sided test are: H 0 : σ 2 = 4 . 8 vs. H 1 : σ 2 > 4 . 8 The test statistic can be calculated as T = ∑ n i =1 ( X i - ¯ X ) 2 σ 2 0 = 133 4 . 8 = 27 . 7 where T ∼ χ 2 n - 1 = χ 2 15 Looking at the chi-squared table, we see that this test statistic of 27 . 7 exceeds the point χ 2 15;0 . 05 = 27 . 488. There is strong evidence to suggest that σ 2 > 4 . 8, and we would reject this null hypothesis at a significant level of at least 2 . 5% Note: We can use software to calculate the exact p-value as Pr( T ≥ 27 . 7) = 0 . 0235 4. Given that n = 21 and s 2 = 5 . 1, we can use the properties of the χ 2 - distribution to construct a 95% confidence interval for σ 2 as follows: Pr " χ 2 20;0 . 975 ≤ ∑ 21 i =1 ( X i - ¯ X ) 2 σ 2 ≤ χ 2 20;0 . 025 # = 0 . 95 Pr " ∑ 21 i =1 ( X i - ¯ X ) 2 χ 2 20;0 . 025 ≤ σ 2 ≤ ∑ 21 i =1 ( X i - ¯ X ) 2 χ 2 20;0 . 975 # = 0 . 95 1

We can now use the fact that ∑ n i =1 ( X i - ¯ X ) 2 = ( n - 1) S 2 and determine the above percentage points: χ 2 20;0 . 025 = 34 . 17 and χ 2 20;0 . 975 = 9 . 591 Yielding Pr (20)(5 . 1) 34 . 17 ≤ σ 2 ≤ (20)(5 . 1) 9 . 591 = 0 . 95 Therefore, a 95% confidence interval for σ 2 is: [2 . 985; 10 . 635] 8. (a) 95% Confidence Interval for μ , assuming σ 2 = 27: Pr ¯ X - z 0 . 025 σ √ n ≤ μ ≤ ¯ X + z 0 . 025 σ √ n = 0 . 95 Now we are given that σ 2 = 27. From the data we see that n = 18 and ¯ X = 80 . 39. From the tables we find that z 0 . 025 = 1 . 96. Therefore a 95% confidence interval for μ is given by " 80 . 39 - 1 . 96 r 27 18 ≤ μ ≤ 80 . 39 + 1 . 96 r 27 18 # or [77 . 99 ; 82 . 79] 95% Confidence Interval for μ , assuming σ 2 is unknown: Pr ¯ X - t 17 , 0 . 025 s √ n ≤ μ ≤ ¯ X + t 17 , 0 . 025 s √ n = 0 . 95 where s 2 is the sample variance. From the data we see that s = 5 . 15 and from the tables we find that t 17 , 0 . 025 = 2 . 11. Therefore a 95% confidence interval for μ is given by 80 . 39 - 2 . 11 5 . 15 √ 18 ≤ μ ≤ 80 . 39 + 2 . 11 5 . 15 √ 18 or [77 . 83 ; 82 . 95] 2

UNIVERSITY OF CAPE TOWN DEPARTMENT OF STATISTICAL SCIENCES STA2030S 2023 Tutorial 8 Extra Question Show all your workings . 1. A random sample X 1 , X 2 , . . . , X n is selected from a normal population with mean μ and standard deviation 1. Later an additional independent observation X n +1 is obtained from the same population. What is the sampling distribution of the statistic ( X n +1 - μ ) 2 + ∑ n i =1 ( X i - X ) 2 , where X denotes the sample mean? 2. Let { 6 . 6 , 5 . 9 , 4 . 4 , 6 . 4 , 7 . 8 , 5 . 2 , 5 . 8 , 5 . 5 , 6 . 4 } be the set of observed values of a random sample of size 9 from a normal distribution with known mean of 6 and variance σ 2 . You are given the statistic ∑ 9 i =1 ( x i - μ ) 2 = 7 . 42. (a) Construct a 95% confidence interval for σ 2 . (b) Test at the 5% signicance level the claim that σ 2 = 0 . 85. 1

Solutions: 1. X n +1 - μ √ 1 2 ∼ N (0 , 1) → ( X n +1 - μ ) 2 ∼ χ 2 1 : by Theorem 3.2. 4 ∑ n i =1 ( X i - X ) 2 √ 1 2 ∼ χ 2 n - 1 : by Theorem 4.2. 4 Thus ( X n +1 - μ ) 2 + ∑ n i =1 ( X i - X ) 2 ∼ χ 2 n : Sum of independent χ 2 random variables. 4 (a) 95% confidence interval for σ 2 : χ 2 9;0 . 025 = 19 . 023 and χ 2 9;0 . 975 = 2 . 700 4 So the 95% onfidence interval for σ 2 is: 7 . 42 19 . 023 ; 7 . 42 2 . 700 = [0 . 3706 ; 2 . 7481] 44 (b) H 0 : σ 2 = 0 . 85 vs H 0 : σ 2 > 0 . 85 4 Test statistic: W T = ∑ 9 i =1 ( x i - μ ) 2 σ 2 0 = 7 . 42 0 . 85 = 8 . 7294 4 Critical value: χ 2 9 , 0 . 05 = 16 . 919 4 Conclusion: Since W T = 8 . 7294 < χ 2 9 , 0 . 05 = 16 . 919, we do not have enough evidence to reject the null hypothesis. 4 2