diff --git a/source/main.ptx b/source/main.ptx index 759ca1a..85332b5 100644 --- a/source/main.ptx +++ b/source/main.ptx @@ -40,6 +40,7 @@ + diff --git a/source/quizzes/quiz-03.ptx b/source/quizzes/quiz-03.ptx new file mode 100644 index 0000000..63081e8 --- /dev/null +++ b/source/quizzes/quiz-03.ptx @@ -0,0 +1,326 @@ + + + + + Quiz 3 + + + +

+ The following work should be completed individually. + Use of notes or textbooks is not allowed. + You may use a scientific calculator, not a graphing calculator or phone app. +

+ +

+ Show all work unless instructed otherwise. +

+ + + + + + + +

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

+ + + + + +

+ Suppose a parameter \theta \in (-\infty, \infty) has likelihood function \mathcal{L}(\theta) with \L'(\theta) = (1 - \theta)e^{\theta}. + Then the maximum likelihood estimation is \widehat{\theta} = 1. +

+ + + +

+ True. +

+ + + + + + +

+ Suppose X_1, \dotsc, X_n are random variables, and S = X_1 + \dotsb + X_n. + Then, for sufficiently large n, S \approx \operatorname{N}(0, 1). +

+ + + +

+ False. +

+ + + + + + +

+ Suppose we collect data to estimate the value of a parameter \theta. + Based on the collected data, we find a 95% confidence interval [a, b] and a 90% confidence interval [c, d]. + Then a \leq c \leq d \leq b. +

+ + + +

+ True. +

+ + + + + + +

+ The heights of five plants are measured and recorded below. + Find the sample mean, sample variance, and a 95% confidence interval around the sample mean for the heights of the plants. + (Pretend that 5 measurements is enough for the CLT to apply.) +

+ + + + Plant i + H_i (in) + + + + 1 + 13 + + + + 2 + 14 + + + + 3 + 10 + + + + 4 + 15 + + + + 5 + 17 + + + + + +

+ Let A = \frac{H_1 + \dotsb + H_5}{5} be the average height of the sample. + Then: + + A \amp = 13.8 \amp s^2 \amp = \frac{\sum H_i^2 - n(A^2)}{n - 1} = \frac{979 - 5(13.8^2)}{4} = 6.7 + \sum H_i^2 \amp = 979 \amp \sqrt{\frac{s^2}{n}} \amp = \sqrt{\frac{6.7}{5}} \approx 1.16 + + So the 95% confidence limits are: + + \mu_{\ell} \amp = 13.8 - 1.96(1.16) \approx \boxed{11.53} + \mu_{h} \amp = 13.8 + 1.96(1.16) \approx \boxed{16.07} + +

+ + + + + + + + +

+ Suppose a coin comes up heads with probability 0.6. + We flip the coin 80 times, and let S be the number of heads. + Estimate \Pr(42 \leq S \leq 50). + (Use a continuity correction if appropriate.) +

+ + + +

+ S \sim \Bin(80, 0.6), so \E(S) = (80)(0.6) = 48 and \Var(S) = (80)(0.6)(0.4) = 19.2. By the CLT, S \approx \Norm(48, 19.2), so: + + \Pr(42 \leq S \leq 50) \amp \approx \Pr\left( \frac{41.5 - 48}{\sqrt{19.2}} \leq Z \leq \frac{50.5 - 48}{\sqrt{19.2}}\right) + \amp \approx \Pr(-1.48 \leq Z \leq 0.57) + \amp = \Phi(0.57) - \Phi(-1.48) + \amp \approx 0.7157 - 0.0694 + \amp = \boxed{0.6463} + +

+ + + + + <m>\Phi(z)</m> Values + + + + z + 0.00 + 0.01 + 0.02 + 0.03 + 0.04 + 0.05 + 0.06 + 0.07 + 0.08 + 0.09 + + + + −1.6 + 0.0548 + 0.0537 + 0.0526 + 0.0516 + 0.0505 + 0.0495 + 0.0485 + 0.0475 + 0.0465 + 0.0455 + + + + −1.5 + 0.0668 + 0.0655 + 0.0643 + 0.0630 + 0.0618 + 0.0606 + 0.0594 + 0.0582 + 0.0571 + 0.0559 + + + + −1.4 + 0.0808 + 0.0793 + 0.0778 + 0.0764 + 0.0749 + 0.0735 + 0.0721 + 0.0708 + 0.0694 + 0.0681 + + + + −1.3 + 0.0968 + 0.0951 + 0.0934 + 0.0918 + 0.0901 + 0.0885 + 0.0869 + 0.0853 + 0.0838 + 0.0823 + + + + −1.2 + 0.1151 + 0.1131 + 0.1112 + 0.1093 + 0.1075 + 0.1056 + 0.1038 + 0.1020 + 0.1003 + 0.0985 + + + + 0.2 + 0.5793 + 0.5832 + 0.5871 + 0.5910 + 0.5948 + 0.5987 + 0.6026 + 0.6064 + 0.6103 + 0.6141 + + + + 0.3 + 0.6179 + 0.6217 + 0.6255 + 0.6293 + 0.6331 + 0.6368 + 0.6406 + 0.6443 + 0.6480 + 0.6517 + + + + 0.4 + 0.6554 + 0.6591 + 0.6628 + 0.6664 + 0.6700 + 0.6736 + 0.6772 + 0.6808 + 0.6844 + 0.6879 + + + + 0.5 + 0.6915 + 0.6950 + 0.6985 + 0.7019 + 0.7054 + 0.7088 + 0.7123 + 0.7157 + 0.7190 + 0.7224 + + + + 0.6 + 0.7257 + 0.7291 + 0.7324 + 0.7357 + 0.7389 + 0.7422 + 0.7454 + 0.7486 + 0.7517 + 0.7549 + + +
+ + \ No newline at end of file