Exam 1 Review ProblemsStatistics & Samples (Ch. 1)Displaying & Describing Data (Chs. 2-3)Probability (Ch. 5)Normal Curve (Ch. 10)Central Limit Theorem (Ch. 4)Hypothesis Testing (Ch. 6)Experimental Design (Ch. 14)1. Harry believes that the Whomping Willow tree grows taller faster when it is exposed to temperatures greater than 25 degrees C. To test this theory, he plants 100 Whomping Willow seeds in large pots. He places a randomly-chosen 50 of them in a greenhouse on the eastern side of Hogwarts and sets the temperature to remain at a constant 30 degrees C. He places the remaining 50 plants in a separate greenhouse on the western side of Hogwarts and sets the temperature to 10 degrees C. a. Is this an observational or experimental study? How do you know?Experimental. Random assignment of plants to treatment (10 degrees of 30 degrees). Manipulation of a variable (temperature).b. The variable height is what type and scale?Numeric, continuousc. What is the explanatory variable? What is the response variable?Exp = temperature; Res = heightd. What are Harry’s null and alternative hypotheses?Ho: There is no height difference in trees grown in 10 degrees and trees grown in 30 degreesHa: Trees grown in 30 degrees will be taller than trees grown in 10 degrees (based on his expectation, this hypothesis needs to be one-sided)e. Harry’s experiment suffers from what bias? How could he reduce this bias? (More than one possible answer.)The trees on the eastern side of the school will be exposed to different sunlight than the ones on the western side. This means that a growth difference could be caused by a 3rd variable (i.e. sunlight) rather than the temperature of the greenhouses. Harry should try blocking—have two greenhouses (one hot, one cold) on both sides of Hogwarts (or on the same side).He isn’t blind to the study. He may water them differently, remove insects differently, etc. because he knows the study’s hypothesis. Have a person unfamiliar with Harry’s hypothesistake care of the plants while they grow.f. At the end of 2 years, Harry records the height of the 50 trees in the hot greenhouse and the height of the 50 trees in the cold greenhouse. What type of graph should Harry use to display this data? (More than one possible answer.)One categorical variable (temperature) and one numeric variable (height)Grouped box plotGrouped histogramGrouped cumulative frequency distributiong. In looking at his data, Harry notices that the heights in both the hot and cold groups are negatively skewed. In fact, he has a few seeds that never grew and he would consider outliers. What should Harry use to report the spread of his data? (Choose one of the following)i) median –not a measure of spreadii) standard deviation --only when data is not skewed and no outliersiii) interquartile range (IQR) –because data is skewed and there are outliersiv) mean –not a measure of spread2. The undergraduate student body at UT is broken down into the following years: 31% are freshmen, 27% are sophomores, 20% are juniors, and 22% are seniors (or higher). 18% of sophomores, 22% of juniors, and 28% of seniors have attended at least one football game. a) If 22% of the entire undergraduate body has attended a football game, what percent of freshmen have gone to a game? Pr[game] = Pr[game I freshman]Pr[freshman] + Pr[game I soph]Pr[soph] + Pr[game I jr]Pr[jr] + Pr[game I sr]Pr[sr].22 = Pr[game I freshman] (.31) + (.18)(.27) + (.22)(.20) + (.28)(.22).22 = Pr[game I freshman] (.31) + .0486 + .044 + .0616.22 = Pr[game I freshman] (.31) + .1542.0658 = .31Pr[game I freshman]Pr[game I freshman] = .2126b) Draw a probability tree of the above information. (Include the number you found in part a.)c) If you find a random student who has never attended a football game, what is the probability that he is a freshman?Pr[fresh I no game] = (Pr[no game I freshman]Pr[freshman]) / Pr[no game]In part a, you learned that 22% of the student body has gone to a game. Thus, (1 - .22), or .78, have not gone to a game.Pr[fresh I no game] = (1-.2126)*(.31) / .78= .31293. The lifespan of the mayfly is normally distributed with a mean of 23.7 hours and a standard deviation of 1.6 hours.a) What percent of mayflies live at least 26.8 hours?z = (26.8 – 23.7) / 1.6z = 1.94 Pr[z < 1.94] = 0.9738Pr[z > 1.94] = 1 – 0.9738 = 0.02622.62%b) 85% of mayflies die after how many hours?Z = 1.041.04 = (x – 23.7) / 1.625.36 hoursc) What is the interquartile range of the lifespan of the mayfly? Q1 = 25th percentile; z = -.67-.67 = (x – 23.7) / 1.6x = 22.628Q3 = 75th percentile; z = .67.67 = (x – 23.7) / 1.6x = 24.772IQR = Q3 – Q1 = 24.772 – 22.628 = 2.144d) In your lab, you take a sample of 5 mayflies and measure their lifespan. You find that theylive an average of 24.5 hours long. What is the probability of that?SE = (1.6) / sqrt (5) = 0.7155z = (24.5 – 23.7) / 0.7155z = 1.12Pr[z < 1.12] = .8686 (This is the probability of finding 5 mayflies that average 24.5 hours orless)1 - .8686 = .1314 = The probability of finding 5 mayflies that average 24.5 hours or moree) You think the results in part d might be some kind of a fluke, so this time you measure the lifespan of 100 mayflies and find the exact same mean value: they live an average of 24.5 hours. What is the probability of a group this size averaging a lifespan of 24.5 hours?SE = 1.6 / sqrt 100 = .16z = (24.5 – 23.7) / .16 = 5pr[z > 5] = really, really, really low (too low to be on your table)f) In part e, you collected data on the lifespan of 100 flies. What is the best way to display this data? (More than one possible answer)single variable, numeric datahistogram, box plot, cumulative frequency diagram4. Two treatments for thyroid cancer are surgery and radioactivity. Martha conducts a studywhere she measures the recovery time of patients receiving one of these two treatments. Martha’s null hypothesis is that the two recovery times do not differ. Her alternative hypothesis is that the radioactivity treatment results in a much quicker recovery than the surgical treatment.a) What is the population for Martha’s study?All individuals with thyroid cancer(if you want to be super specific, it would be all individuals with thyroid cancer who are choosing either surgery or radioactivity)b) Martha would like to use a random sample, but she has to use a volunteer sample instead. What is a volunteer sample, and why does she have to use it?Volunteer sample—individuals who volunteer to