Stat 204 Sample Exam 1 Name:

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1 Stat 204 Sample Exam 1 Name: 1. (Peck, 1.75, p.214) Consider the following study: Two hundred people were randomly selected from a list of all people living in Minneapolis who receive Social Security. Each person in the sample was asked whether or not they took calcium supplements. These people were followed for five years, and whether or not they had had a heart attack during the five year period was noted. The proportion of heart attack victims in the group taking calcium supplements was significantly higher than the proportion of heart attack victims in the group not taking calcium supplements. a. Is the study described an observational study or an experiment? This is an observational study because events, and not the researchers, placed the participants into the two groups, had a heart attack within five years or not. b. Did the study use random selection from some population? Participants were randomly selected from a list of all people living in Minneapolis who receive Social Security. c. Did the study use random assignment to experimental groups? The study did not use random assignment to experimental groups. d. Based on the study description, would it be reasonable to conclude that taking calcium supplements is the cause of the increased risk of heart attack? The study is an observational study, so we cannot use it to conclude causality. e. Would it be reasonable to generalize conclusions from this study to some larger population? If so, what population? Because random selection from the list of all people living in Minneapolis who receive Social Security was used to assemble the participants, the conclusions from the study can be generalized to the set of all people living in Minneapolis who receive Social Security. 1

2 2. (Peck, 3.69, p.152) Sugar content (in grams per serving) for 18 high-fiber cereals are shown. Create a dotplot Sugar content (grams per serving) Calculate the five-number summary and IQR for this data. summary(sugar) ## Min. 1st Qu. Median Mean 3rd Qu. Max. ## IQR(sugar) ## [1] 4.5 Create a boxplot Sugar content (grams per serving) a. Calculate the lower boundary, bdy.lo, with the property that any data value smaller than bdy.lo is an outlier. 2

3 q1 <- as.numeric(quantile(sugar, 0.25)) iqr <- IQR(sugar) bdy.lo <- q1-1.5 * iqr bdy.lo ## [1] 0.75 b. Here are the smallest 5 values in the data set. Which of these values are outliers? head(sort(sugar), 5) ## [1] c. Calculate the upper boundary, bdy.hi, with the property that any data value larger than bdy.hi is an outlier. q3 <- as.numeric(quantile(sugar, 0.75)) iqr <- IQR(sugar) bdy.hi <- q * iqr bdy.hi ## [1] d. Here are the largest 5 values in the data set. Which of these values are outliers? tail(sort(sugar), 5) ## [1]

4 3. (Peck, 4.66, p.214) Studies have shown that people who suffer sudden cardiac arrest have a better chance of survival if a defibrillator shock is administered very soon after cardiac arrest. How is survival rate related to the length of time between cardiac arrest and the defibrillator shock being delivered? The accompanying data give y = survival rate (percent) and x = mean call-to-shock time (minutes) for a cardiac rehabilitation center (in which cardiac arrests occurred while victims were hospitalized and so the call-to-shock time tended to be short) and for four other communities of different sizes. Mean call-to-shock time, x Survival rate, y Display the data. Survival rate (%) Mean call to shock time (min) a. Find the equation of the least squares line. defibrillator.lm$coefficients ŷ = x ## (Intercept) x ## b. Interpret the slope of the least squares line in the context of this study. If the mean call-to-shock time increases by 1 minute, then the survival rate will drop by c. Does it make sense to interpret the intercept of the least squares regression line? If so, give an interpretation. If not, explain why it is not appropriate for this data set. 4

5 The mean call-to-shock time cannot be 0, so we do not interpret the intercept of the least squares regression line. d. Use the least squares line to predict survival rate for a community with a mean call-to-shock time of 10 minutes. new.data <- data.frame(x = 10) y.hat <- predict(defibrillator.lm, new.data) y.hat ## 1 ## Add this point to the chart. plot(x, y, pch=20, col="darkred", xlab="mean call-to-shock time (min)", ylab="survival rate (%)") defibrillator.lm <- lm(y ~ x) abline(defibrillator.lm, col="orange") points(x=10, y=y.hat, pch=20, col="green") Survival rate (%) Mean call to shock time (min) 5

6 4. (Peck, 5.84, p.272) Is ultrasound a reliable method for determining the gender of an unborn baby? Consider the following data on 1,000 births, which are consistent with summary values that appeared in an online journal. ## predicted.gender ## gender predicted f predicted m Sum ## f ## m ## Sum Illustration. mosaicplot(t(ultrasound), col=c("green", "yellow")) t(ultrasound) predicted f predicted m m gender f predicted.gender Calculate the following probabilities. a. P (f) 480 / 1000 ## [1] 0.48 b. P (predicted f) 6

7 562 / 1000 ## [1] c. P (f predicted f) 432 / 562 ## [1] d. P (predicted f f) 432 / 480 ## [1] 0.9 7

8 5. (Peck, 6.59, p.314) The amount of time spent by a statistical consultant with a client at their first meeting is a random variable that has a normal distribution with a mean value of 60 minutes and a standard deviation of 10 minutes. Let X be a random variable returning the time spent by the statistical consultant and client at their first meeting, X N(µ = 60, σ = 10) Illustration. curve(dnorm(x, mean=60, sd=10), from=30, to=90, las=1, col="darkred", xlab="time (min)", ylab="density", main=expression(x %~% N(mu, sigma))) X ~ N(µ, σ) Density Time (min) a. What is the probability that more than 45 minutes is spent at the first meeting? mu <- 60 sigma <- 10 x <- 45 z <- (x - mu) / sigma z ## [1]

9 p.value <- 1 - pnorm(z) p.value ## [1] b. What amount of time is exceeded by only 10% of all clients at a first meeting? z.star <- qnorm(0.90) x.star <- mu + z.star * sigma x.star ## [1]

10 6. (Peck, 6.75, p.327) Suppose that in a certain metropolitan area, 90% of all households have cable TV. Let x denote the number among four randomly selected households that have cable TV. Then x is a binomial random variable with n = 4 and p = 0.9. Let the random variable X have a binomial distribution with n = 4 and p = 0.9, Distribution. X Binomial(n = 4, p = 0.9). X ~ Binomial(n=4, p=0.9) probability k Calculate the following probabilities. a. P (X = 4) dbinom(4, n, p) ## [1] b. P (1 X 3) sum(dbinom(1:3, n, p)) ## [1] c. P (X = 2) 10

11 dbinom(2, n, p) ## [1] d. P (X 2) 1 - dbinom(2, n, p) ## [1]