hw4_sol

Stats 120B Homework 4: Solution 1. Assume X 1 , . . . , X n form a random sample from N ( µ, σ 2 ) with both µ and σ 2 unknown. (a) What is the sampling distribution of ¯ X ? State the result. You don’t have to prove it. ¯ X ∼ N ( µ, σ 2 /n ) (b) State a result about the distribution of an expression involving the sample variance S 2 = ∑ n i =1 ( X i − ¯ X ) 2 / ( n − 1). You don’ have to prove it. ( n − 1) S 2 σ 2 ∼ χ 2 ( n − 1) Note that any other results that involves S 2 is also fine. (c) Is T a statistic? Is T a pivotal quantity? Explain. Under H 0 , T is a statistic because there is no parmater in it ( µ 0 is a fixed constant). T is a pivotal quantity since its distribution does not have any parameters. (d) Assume ( − 1 , 2) is a 90% confidence interval for µ , find a 90% confidence in- terval for 1 /µ . P ( − 1 < µ < 2) = . 90 ⇒ P ( µ − 1 > 1 / 2or µ − 1 < − 1) = . 90 So CI is ( −∞ , − 1) ∪ (1 / 2 , ∞ ). Read the following R code and output, answer questions (e)–(j). > n = length(x) > n [1] 100 > mean(x) [1] 1.49067 > var(x) [1] 15.95129 > stat = (mean(x) - 0)/(sqrt(var(x)/(n))) > stat [1] 3.732360 > pt(stat,n-1,lower=F) *2 [1] 0.0003167425 (e) Find the value of n , ¯ X , S 2 and µ 0 . 1

n = 100 , ¯ X = 1 . 49 , S 2 = 15 . 95 , µ 0 = 0 (f) State the alternative hypothesis. H a : µ ̸ = 0 (g) State your conclusion of hypothesis test at the significance level of α = . 05. p-value is 0 . 0003167, so we reject the null hypothesis at significance level of α = . 05. There is enough evidence to reject µ = 0. (h) State your conclusion of hypothesis test at the significance level of α = . 0001. p-value is above α now. So we don’t have enough evidence against the null hypothesis. (i) What are the assumptions we make on the data? (1) normality; (2) independence; (3) identically distributed. (j) Here is the QQ plot of the data. Which assumption is checked here? Does the data satisfy that assumption? -3 -2 -1 0 1 2 3 -0.2 0.0 0.2 0.4 0.6 Theoretical Quantiles Sample Quantiles QQ plot is used to check normality assumption. And it appears that normality is not satisfied for the data. (The answer here is subjective, you can say that the normality assumption looks reasonable if you think the curve looks close to the straight line. ) 2

(g) The p -value of a hypothesis test is the probability that the null hypothesis is correct. False. Again, we cannot talk about the probability of a null hypothesis. The p -value is the probability of the data (or something more extreme), given that the null hypothesis is true. (h) If a test rejects the null hypothesis at a significance level of 0.06, then the p -value is less than or equal to 0.06. True. A significance level of 0.06 implies that we reject the null when the p -value is less than or equal to 0.06 (otherwise we fail to reject the null). 3. An airline suspects that a majority of frequent fliers prefer aisle seats. They define p = proportion of the population of frequent fliers who prefer aisle seats, and plan to test the hypotheses H 0 : p = 0 . 5 versus H a : p > 0 . 5. The company statistician plans to conduct the test using a significance level of 0.05 based on a random sample of 600 frequent fliers. Suppose in fact 55% of the population of frequent fliers prefer aisle seats. Then the probability that the test will (correctly) result in rejecting the null hypothesis is 0.791. Based on this information, provide numerical values for each of the following. (a) the null set, Ω 0 Ω 0 = { 0 . 5 } (b) the power of the test 0.791 (c) the probability of making a Type 1 error (think carefully about this one!) The probability of making a Type 1 error = 0 (since the null hypothesis is false) (d) the probability of making a Type 2 error The probability of making a Type 2 error = 1 − 0 . 791 = 0.209 4. Suppose that n light bulbs are burning simultaneously to determine the lengths of their lives. We shall assume that they burn independently and that the lifetime of each bulb has the exponential distribution with parameter β . Let X i denote the lifetime (in thousand hours) of bulb i , for i = 1 , . . . , n . The pdf, mean and variance 4

of exponential distribution Exp( β ) are f ( x ) = βe − βx , E( X ) = 1 β , Var( X ) = 1 β 2 . (a) What does the central limit theorem say about the distribution of √ n ( ¯ X − 1 /β ) 1 /β when n is large? Its approximation distribution is N (0 , 1) when n is large. (b) Find a two-sided 95% confidence interval of the mean lifetime (1 /β ). Express your answer in terms of n , ¯ X and z . Interpret the confidence interval you obtain in context of the situation. Use probability equation P ( z . 025 ≤ √ n ( ¯ X − 1 /β ) 1 /β ≤ z . 975 ) = . 95 ⇒ P ( ¯ X z . 975 / √ n + 1 ≤ 1 β ≤ ¯ X z . 025 / √ n + 1 ) = . 95 So the 95% CI is ( ¯ X z . 975 / √ n +1 , ¯ X z . 025 / √ n +1 ). We are 95% confident that the mean lifetime falls in this interval. Another solution is to use an estimator (say MLE) for β , denoted by ˆ β , then use P ( z . 025 ≤ √ n ( ¯ X − 1 /β ) 1 / ˆ β ≤ z . 975 ) = . 95 ⇒ P ( ¯ X − z . 975 √ n ˆ β ≤ 1 β ≤ ¯ X − z . 025 √ n ˆ β ) = . 95 So 95% CI is ( ¯ X − z . 975 √ n ˆ β , ¯ X − z . 025 √ n ˆ β ) (c) Now, suppose that we want to test whether or not the mean lifetime is 1000 hours, we would consider H 0 : β = 1 ←→ H a : β ̸ = 1 . Derive the likelihood ratio statistic. The likelihood ratio statistic is Λ = max β : β =1 L ( β | X ) max L ( β | X ) = L ( β = 1 | X ) L ( β = ¯ X − 1 | X ) = exp( − ∑ X i ) 1 ¯ X n exp( − 1 ¯ X ∑ X i ) = ¯ X n exp( − n ¯ X + n ) . 5