HWK5

Stat 371 Homework #5 Rachel Marten • Submit your homework to Canvas by the due date and time. Email your lecturer if you have extenuating circumstances and need to request an extension. • If an exercise asks you to use R, include a copy of all relevant code and output in your submitted homework file. You can copy/paste your code, take screenshots, or compile your work in an Rmarkdown document. • If a problem does not specify how to compute the answer, you many use any appropriate method. I may ask you to use R or use manual calculations on your exams, so practice accordingly. • You must include an explanation and/or intermediate calculations for an exercise to be complete. • Be sure to submit the HWK5 Autograde Quiz which will give you ~20 of your 40 accuracy points. • 50 points total: 40 points accuracy, and 10 points completion Sampling Distributions and CLT Exercise 1. Retail stores experience their heaviest volume of transactions that include returns on December 26th and December 27th each year. The distribution for the Number of Items Returned (X) by Macy’s customers who do a return transaction on those days last year is given in the table below. It has mean: μ = 2.61 and variance σ 2 ≈ 1.80 . Number of Items Returned in Transaction (x) Probability 1 0.25 2 0.28 3 0.20 4 0.17 5 0.08 6 0.02 vals <- c ( 1 , 2 , 3 , 4 , 5 , 6 ) probs <- c ( 0.25 , 0.28 , 0.20 , 0.17 , 0.08 , 0.02 ) EV_pop = sum (vals * probs) Var_pop <- sum (probs * (vals - EV_pop) ^ 2 ) a. Is this population distribution left skewed, symmetric, or right skewed? How do you know? This population is right skewed sine there is more data closer to 1-4. This

makes it heavy on the left side of the graph leaving a tail on the right part. The mean is 2.61 which is on the lower side of the data and the standard deviation also demonstrates a right skewed distribution. b. What proportion of returns had three or more items? 47% of returns had three or more items. 0.20+0.17+0.08+0.02=0.47 1 - pnorm ( 3 , 2.61 , sqrt ( 81 )) ## [1] 0.4827179 c. Identify which histogram below diplays (1) the population X values, (2) the simulated sampling distribution of the sample mean ´ X , (3) the simulated sampling distribution of the sample total T . Briefly explain how you know. c. Histogram A corresponds to the simulated sampling distribution of the sample total. Histogram A shows the mean to be around 117.45 which is the sample mean. Histogram B corresponds to the population x values. This histogram has all fo the numbers of items returned with the approximations of each. Histogram C corresponds to the simulated sampling distribution of the sample mean. The mean of histogram C is the same as the sample mean data. d. Describe the sampling distribution (shape, mean, and standard deviation) of the sample mean number of items returned in 45 return transactions

a. Construct histograms and qqnorm plots for all three of the quantitative variables recorded on the 31 trees. For which of the three variables do we have the strongest evidence that the population of values may not be well approximated by a normal random variable? volumes <- trees $ Volume (x_bar <- mean (volumes)) ## [1] 30.17097 (s <- sd (volumes)) ## [1] 16.43785 (n <- length (volumes)) ## [1] 31 par ( mfrow = c ( 1 , 2 )) hist (volumes) qqnorm (volumes); qqline (volumes) Girth <- trees $ Girth (x_bar <- mean (Girth))

## [1] 13.24839 (s <- sd (Girth)) ## [1] 3.138139 (n <- length (Girth)) ## [1] 31 par ( mfrow = c ( 1 , 2 )) hist (Girth) qqnorm (Girth); qqline (Girth) Height <- trees $ Height (x_bar <- mean (Height)) ## [1] 76 (s <- sd (Height)) ## [1] 6.371813 (n <- length (Height)) ## [1] 31

## [1] 6.371813 ## [1] 31 mean (trees $ Volume); sd (trees $ Volume); length (trees $ Volume) ## [1] 30.17097 ## [1] 16.43785 ## [1] 31 qt (. 9 , 30 ) ## [1] 1.310415 # standard error se <- 3.14 / sqrt ( 31 ) # critical value t <- qt ( 0.05 , df = 30 ) # margin of error moe <- 1.697261 * 0.5636 1.697261 * 0.5636 ## [1] 0.9565763 #upper bound 13.25+0.9565763 ## [1] 14.20658 #lower bound 13.25-0.9565763 ## [1] 12.29342 #height 6.371813 / sqrt ( 31 ) #se ## [1] 1.144411 qt ( 0.95 , 30 ) #criticalvalue ## [1] 1.697261 1.697261 * 1.144411 #marginoferror ## [1] 1.942364 76 +1.942364 #upper bound

## [1] 77.94236 76 -1.942364 #lower bound ## [1] 74.05764 #girth 3.138139 / sqrt ( 31 ) #se ## [1] 0.5636264 qt ( 0.95 , 30 ) #critical value ## [1] 1.697261 1.697261 * 0.5636 #margin of error ## [1] 0.9565763 13.25+0.9565763 #upper bound ## [1] 14.20658 13.25-0.956763 #lower bound ## [1] 12.29324 #volume 16.43785 / sqrt ( 31 ) #se ## [1] 2.952325 qt ( 0.95 , 30 ) #criticalvalue ## [1] 1.697261 1.697261 * 2.95 #marginoferror ## [1] 5.00692 30.17+5.00692 #upper bound ## [1] 35.17692 30.17-5.00692 #lower bound ## [1] 25.16308 c. Construct the same confidence intervals that you constructed in (b) above using the t.test() command in R. Confirm that you get very similar endpoints. se <- 3.14 t.test (trees, conf.level = 0.9 )