<>= n <- sample(8:15, 1) y <- rnorm(n, runif(1, 4940, 4990), runif(1, 30, 50)) alpha <- sample(c(0.1, 0.05, 0.01), 1) Mean <- round(mean(y), digits = 1) Var <- round(var(y), digits = 1) sd <- sqrt(Var/n) fact <- round(qt(1 - alpha/2, df = n - 1), digits = 4) facn <- round(qnorm(1 - alpha/2), digits = 4) LBt <- round(Mean - fact * sd, digits = 3) UBt <- round(Mean + fact * sd, digits = 3) LBn <- round(Mean - facn * sd, digits = 3) UBn <- round(Mean + facn * sd, digits = 3) ## use extended Moodle processing to award 100% for correct solution based on ## t quantiles and 50% for the solution based on normal quantiles ## ## this can be handled as a "verbatim" solution, directly including Moodles ## cloze type: ## ":NUMERICAL:=2.228:0.1~%50%1.960:0.1#Approximately normal instead of exact t solution." ## where 2.228 is the correct and 1.960 the partially correct solution, ## the tolerance is 0.01 in both cases, and a comment is supplied at the end. ## More details: https://docs.moodle.org/35/en/Embedded_Answers_(Cloze)_question_type ## solution template (note: % have to be escaped as %% for sprintf) sol <- ":NUMERICAL:=%s:0.1~%%50%%%s:0.1#Approximately normal instead of exact t solution." ## insert correct and partially correct solutions sol <- sprintf(sol, c(LBt, UBt), c(LBn, UBn)) @ \begin{question} It is suspected that a supplier systematically underfills 5~l canisters of detergent. The filled volumes are assumed to be normally distributed. A small sample of $\Sexpr{n}$ canisters is measured exactly. This shows that the canisters contain on average $\Sexpr{Mean}$~ml. The sample variance $s^2_{n-1}$ is equal to $\Sexpr{Var}$. Determine a $\Sexpr{100 * (1 - alpha)}\%$ confidence interval for the average content of a canister (in ml). \begin{answerlist} \item What is the lower confidence bound? \item What is the upper confidence bound? \end{answerlist} \end{question} \begin{solution} The $\Sexpr{100 * (1 - alpha)}\%$ confidence interval for the average content $\mu$ in ml is given by: \begin{eqnarray*} & & \left[\bar{y} \, - \, t_{n-1;\Sexpr{1-alpha/2}}\sqrt{\frac{s_{n-1}^2}{n}}, \; \bar{y} \, + \, t_{n-1;\Sexpr{1-alpha/2}}\sqrt{\frac{s_{n-1}^2}{n}}\right] \\ & = & \left[ \Sexpr{Mean} \, - \, \Sexpr{fact}\sqrt{\frac{\Sexpr{Var}}{\Sexpr{n}}}, \; \Sexpr{Mean} \, + \, \Sexpr{fact}\sqrt{\frac{\Sexpr{Var}}{\Sexpr{n}}}\right] \\ & = & \left[\Sexpr{LBt}, \, \Sexpr{UBt}\right]. \end{eqnarray*} \begin{answerlist} \item The lower confidence bound is $\Sexpr{LBt}$. \item The upper confidence bound is $\Sexpr{UBt}$. \end{answerlist} \end{solution} %% META-INFORMATION %% \extype{cloze} %% \exclozetype{verbatim|verbatim} %% \exsolution{\Sexpr{sol[1]}|\Sexpr{sol[2]}} %% \exname{Confidence interval} %% \extol{0.01}