diff --git a/source/exams/exam-02.ptx b/source/exams/exam-02.ptx new file mode 100644 index 0000000..4f81cf7 --- /dev/null +++ b/source/exams/exam-02.ptx @@ -0,0 +1,387 @@ + + + + + Exam 2 + + + +

+ Show all relevant work. +

+ + + + + + + +

+ Write either True or False for each of the following statements. + No justification is required. +

+ + + + + +

+ Suppose X_1, \dotsc, X_n are any random variables and S = (X_1 + X_2 + X_3 + \dotsb + X_n)^2. + Then S is approximately normally distributed. +

+ + + +

+ False. +

+ + + + + + +

+ Let Z be the standard normal distribution. + For any x, \Pr(Z \geq x) = 1 - \Phi(x). +

+ + + +

+ True. +

+ + + + + + +

+ If we're using a set of data to find a confidence interval for a parameter, then the 95% confidence interval will be wider than the 90% confidence interval. +

+ + + +

+ True. +

+ + + + + + +

+ In hypothesis testing, a type I error is the probability of accepting the null hypothesis. +

+ + + +

+ False. +

+ + + + + + +

+ For a particular set of collected data, the p-value of a 2-tailed test will be as large or larger than the p-value of a 1-tailed test. +

+ + + +

+ True. +

+ + + + + + +

+ In a \chi^2 test, if \chi^2 \lt 0.05, then we can reject the null hypothesis. +

+ + + +

+ False. +

+ + + + + + + + + +

+ In each of the following scenarios, determine whether we should use a 1-tailed test or a 2-tailed test. + Indicate clearly which words or phrase in the description of the scenario would lead you to draw your conclusion. +

+ + + + + +

+ We wonder whether the population of mice in a city have a different average weight to the population of mice in a rural town. +

+ + + +

+ 2-tailed, because we wonder whether the cities have "different average weight" without suspecting a direction of difference. +

+ + + + + + +

+ A new fertilizer is designed to increase the yield of a particular species of fruit, and we wonder whether the fertilizer is effective. +

+ + + +

+ 1-tailed, because the new fertilizer is "designed to increase the yield", so we suspect the change happens in a particular direction. +

+ + + + + + +

+ We find a weighted 6-sided die and suspect that it will roll a 6 more often than a fair die would. +

+ + + +

+ 1-tailed, because we suspect the die will roll 6 "more often", indicating difference in a particular direction. +

+ + + + + + + + + +

+ A coin has a probability \theta = 0.7 of coming up heads. + We flip the coin 180 times, and write S for the number of heads. + Estimate \Pr(120 \leq S \leq 130). + (Use a continuity correction if appropriate.) +

+ + + +

+ S is binomially distributed with E(S) = (180)(0.7) = 126 and \Var(S) = (180)(0.7)(0.3) = 37.8. Therefore: + + \Pr(120 \leq S \leq 130) \amp \approx \Pr\left( \frac{119.5 - 126}{\sqrt{37.8}} \leq Z \leq \frac{130.5 - 126}{\sqrt{37.8}}\right) + \amp \approx \Pr(-1.06 \leq Z \leq 0.73) + \amp = \Phi(0.73) - \Phi(-1.06) + \amp = \Phi(0.73) - \Phi(-1.06) + \amp \approx 0.7673 - 0.1446 + \amp = 0.6227. + +

+ + + + + +

+ The weights of five apples are measured and recorded below. + Find the sample mean, sample variance, and a 95\% confidence interval around the sample mean for the weights of the apples. + (Pretend that 5 measurements is enough for the CLT to apply. + Do not use the t-distribution.) +

+ + + + Apple i + 1 + 2 + 3 + 4 + 5 + + + + Weight W_i (g) + 140 + 120 + 165 + 168 + 137 + + + + + +

+ + A = \text{Avg weight} \amp = \frac{\sum W_i}{5} = 146 + \sum W_i^2 \amp = 108218 + s^2 \amp = \frac{\sum W_i^2 - n(A^2)}{n -1} = \frac{108218 - 5(146^2)}{4} = 409.5 + \sqrt{\frac{s^2}{n}} \amp = \sqrt{\frac{409.5}{5}} \approx 9.05 + \mu_{\ell} \amp = 146 - 1.96(9.05) = 128.262 + \mu_{h} \amp = 146 + 1.96(9.05) = 163.738 + +

+ + + + + + + + +

+ Suppose we're told that a proportion of \theta = 0.4 of a population has a particular trait, but we suspect that this information is incorrect. + We sample 160 people from the population and see 76 people with the trait. + Is this enough evidence to reject the null hypothesis at the 0.05 significance level? +

+ + + +

+ "We suspect the information is incorrect" indicates a 2-tailed test. + Let S be the number of people in the sample with the trait. + Then, under H_0, S \sim \Bin(160, 0.4), so \E(S) = (160)(0.4) = 64 and \Var(S) = (160)(0.4)(0.6) = 38.4. + The observed 76 people is 12 more than expected. + Equally extreme in the other direction would be 12 fewer, so 52 people. + Therefore the p-value is: + + \Pr(S \geq 76) + \Pr(S \leq 52) \amp \approx \Pr\left( Z \geq \frac{75.5 - 64}{\sqrt{38.4}} \right) + \Pr\left(Z \leq \frac{52.5 - 64}{\sqrt{38.4}}\right) + \amp \approx \Pr(Z \geq 1.86) + \Pr(Z \leq -1.86) + \amp = 2 \Phi(-1.86) + \amp = 2 (0.0314) + \amp = 0.0628 \gt 0.05, + + so we do not reject H_0. +

+ + + + + + + + +

+ Suppose we have a coin that we're told has probability \theta = 0.4 of coming up heads, but we suspect it comes up heads more often than that. + We flip the coin 160 times. +

+ + + + + +

+ What is the minimum number of heads we would need to see to reject the null hypothesis? +

+ + + +

+ "More often" suggests we should use a 1-tailed test. + Let S be the number of heads. + Then, under H_0, S \sim \Bin(160, 0.4), so \E(S) = (160)(0.4) = 64 and \Var(S) = (160)(0.4)(0.6) = 38.4. + Therefore, the p-value when observing k heads would be: + + \Pr(S \geq k) \amp \approx \Pr\left( Z \geq \underbrace{\frac{k - 0.5 - 64}{\sqrt{38.4}}}_{z} \right) + \Pr(Z \geq z) \amp = 0.05 + 1 - \Phi(z) \amp = 0.05 \quad \Rightarrow \quad \Phi(z) = 0.95. + + We would need the z-score to be at least 1.65 to reject H_0, so: + + \frac{k - 0.5 - 64}{\sqrt{38.4}} \amp = 1.65 + k = 64.5 + 1.65 \sqrt{38.4} \amp \approx 74.72 + + Therefore, it would take at least 75 heads to reject H_0. +

+ + + + + + +

+ If the coin is actually fair, what is the power of this test? +

+ + + +

+ If the coin is fair, then \E(S) = (160)(0.5) = 80 and \Var(S) = (160)(0.5)(0.5) = 40. + The probability of rejecting H_0 is then: + + \Pr(S \geq 75) \amp \approx \Pr\left(Z \geq \frac{75 - 80}{\sqrt{40}}\right) + \amp \approx \Pr(Z \geq -0.79) + \amp \approx 1 - \Phi(-0.79) + \amp \approx 1 - 0.2148 + \amp = 0.7852. + +

+ + + + + + + + + +

+ A store owner assumes an equal number of people on average come into the store each day of the week. + Are the following counts of 400 observed shoppers consistent with this hypothesis? +

+ + + + Day + Mon + Tue + Wed + Thu + Fri + + + + Observed Shoppers + 83 + 61 + 65 + 90 + 101 + + + + + +

+ Under the null hypothesis, we would expect 400/5 = 80 people each day. + So: + + \chi^2 \amp = \frac{(83 - 80)^2}{80} + \dotsb + \frac{(101 - 80)^2}{80} = 14.2. + + The number of degrees of freedom is (5 - 1) - (0) = 4, so the critical value is 9.488. + Since 14.2 \gt 9.488, we reject the null hypothesis. +

+ + + + \ No newline at end of file diff --git a/source/main.ptx b/source/main.ptx index 10f5a8d..f3b346a 100644 --- a/source/main.ptx +++ b/source/main.ptx @@ -49,6 +49,7 @@ Exams +