Business Statistics Fall Quarter 2015 Midterm Answer Key

Size: px

Start display at page:

Download "Business Statistics Fall Quarter 2015 Midterm Answer Key"

Pearl Ferguson
5 years ago
Views:

A simple calculator may be helpful for computing numerical answers.

1 Business Statistics 4000 Fall Quarter 205 Midterm Answer Key Name (print please): Open book (but be courteous to your neighbors). No internet. A simple calculator may be helpful for computing numerical answers. You have 3 hours to complete the exam. By signing below I attest that I have not violated the Chicago Booth Honor Code either in preparation for, or during, this examination. Signature:

2 . Previous research reports that 25% of young people check Facebook more than twenty times a day. An on-campus survey asks 50 college kids if they check Facebook more than twenty times a day. (a) Assuming that the previous research is accurate for the campus population, describe the outcome of this survey as a binomial random variable. Let the outcome be Y. It is a Binomial random variable Y Bin(50, 0.25). (b) Find the best approximating normal distribution to this binomial random variable. A Binomial random variable can be approximated with a normal distribution N(Np, Np( p)). So in this case, the random variable Y is approximated with the normal distribution N(37.5, 28.25). (c) According to this normal approximation, if this survey were conducted repeatedly, 95% of the time the proportion of people answering yes to the question would be in what range (give an upper and lower number)? The range should be between 37.5 (.96) = 27. and 37.5+(.96) =

3 2. The table below records the probability of different amounts of snowfall. x 0in in 2in 3in 4in 5in 6in 7in 8in 9in 0+in P (X = x) (a) What is the mean amount of snowfall? The mean is (0.07) + 2(0.7) + 3(0.5) + 4(0.5) + 5(0.5) + 6(0.0) + 7(0.05) + 8(0.03) + 9(0.0) + 0(0.02) = Note: I use 0 instead of 0+. (b) What is the modal amount of snowfall? The mode is 2in. (c) What is the median amount of snowfall? The median is 4in. (d) What is the probability of getting between 3 and 6 inches (inclusive) of snow? The probability is

4 3. A local furniture chain holds a big promotion leading up to major holidays and events such as the Super Bowl. The gimmick is that if it snows a certain amount on the day of the event, then anyone who bought furniture during the promotion gets a full refund. From past experience, assume that running the promotion leads to an increase in sales of $ million dollars, for a total sales figure of $3 million. (So they bring in $2 million without the promotion and $3 million with it). On the other hand, if they have to make good on the refunds, they forfeit all sales. (a) Let p denote the probability that they do not have to refund their customers. What value of p makes the promotion worth it, on average? The break-even probability is: 3p + 0( p) = 2, meaning that the expected sales under the promotion scheme should be equal to the sales without such promotion. Solving it gives p = 2. 3 (b) Based on your answer above, use the table from the previous problem to determine how many inches the furniture store should set the promotion threshold at. You may assume that the threshold level does not impact the increase in sales due to participation. The probability of not refunding their customers is equal to the probability of snow amount being smaller than the threshold. The firm wants to the probability of not refunding to be high. From the table, the probability of snow amount being smaller than or equal to 4in (5in) is 0.64 (0.79). So the threshold should be set at least 5in. 4

5 4. Consider two dice, one typical fair six-sided die, with numbers through 6, and one atypical fair three-sided die with numbers 2, 4 and 6. Call the outcome of the six-sided die X and the outcome of the three-sided die Y. Suppose they are rolled independently. (a) What is the probability that X > Y? Let D = X Y. discrete distribution. The distribution of D can be described with the following D p (b) What is the expected value of the difference between X and Y? The expected value is Consider two mutual funds, with historical returns represent by the random variables X N(5, 9) and Y N(8, 4). Let W = X + 2 Y be a composite fund. 3 3 (a) What is the expected value of W? The expected value of W is 3 E(X) E(Y ) = = 3 3 = (b) Assume cor(x, Y ) = 0.3. What is the variance of W? The variance of W is Var(X) + 4Var(Y ) + 2(2)(3) 2 cor(x, Y ) = (c) Give a range for the returns of W that contains 99.7% of the probability mass. Three standard deviation around the mean covers 99.7% of the probability mass, so the range is ( , ] = (6., 4.55). 5

6 6. The overall probability of a pregnancy miscarriage is 0.5. The probability of another miscarriage, given a first miscarriage, is higher. To see why, split the population of pregnant women into two groups, a high risk group and a low risk group. Suppose that the rate of miscarriage among high risk individuals is 0.80 and that high risk individuals comprise 2% of all pregnancies. Let M denote a miscarriage and L denote a low risk pregnancy; let not-l denote a high risk pregnancy. (a) Find the probability of a miscarriage, given that you are a low risk individual. To figure this out, write the overall probability of miscarriage using the law of total probability; then solve for P (M L). The LTP gives P (M) = P (M not-l)p (not-l) + P (M L)P (L) which we can rearrange as P (M L) = P (M) P (M not-l)p (notl) P (L). Plugging in our known values gives ( (0.02)) 0.98 so P (M L) = = 0.4. (b) Use Bayes rule to calculate the probability of being a high risk individual, given that you had a first miscarriage. P (not-l M) = P (M not-l)p (not-l) P (M) = (0.8)(0.02) 0.5 = 0.. (c) Now that you know P (not-l M) (from above), apply the law of total probability again to determine the probability of a second miscarriage. You may assume that, conditional on mother type (high or low risk), subsequent miscarriages are independent. Let M be the second miscarriage and M be the first miscarriage. We write the law of total probability P (M M) = P (M L, M)P (L M) + P (M not-l, M)P (not-l M) = P (M L)P (L M) + P (M not-l)p (not-l M) = 0.4( 0.) + 0.8(0.) = 0.2. The second line follows from the assumption of conditional independence. The final line we plug in the numbers found in the previous two parts. 6

7 7. The hook-up and dating app Tinder asks users to vote yea or nay on prospective partners. After running the app for an hour, you collect 78 responses, of which 24 are yea. Assume that users are served to each other at random; a user must vote in order to see a new face. Give a 95% confidence interval for the proportion of Tinder users in your area who think you d be a suitable match for them. Denote ˆp the estimate of the proportion of users that are suitable match. Given the number of positive reponses, ˆp = 0.3. Using normal distribution N(ˆp, ˆp( ˆp)/78), the 95% CI is [0.085, 0.85]. Is this the distribution correct? 8. On average, do Booth MBA students gain weight during the Winter quarter? Based on a random sample of 3 students, you find that they gained an average of 3.3 pounds, with a standard deviation of 7.73 pounds. Is this amount of weight gain statistically significant at the 0% level? Denote the average weight change by. Under the null hypothesis, has normal distribution N(0, σ 2 /N), where N = 3 is the sample size and σ 2 = is the sample variance. Is this the distribution correct? 3.3 The Z statistics is 7.73/ =.54. One cannot reject the null that there is no weight 3 change on average. The critical value for the alternative hypothesis that the amount of weight gain is not zero is.64. The critical value for the alternative hypothesis that the amount of weight gain is larger than zero is.28. 7

8 9. Fifteen out of 23 children responded favorably to a new allergy treatment, while only 8 out of 50 adults did. Is the efficacy of the drug statistically different between adults and children at the 5% level? Denote the average response by children X and by adults Y. We would like to test whether the difference = X Y is significantly different from zero. Under the null of no difference, N(0, σ 2 ), where we approximate the variance using the observed proportions σ 2 ˆσ 2 = ˆp X( ˆp X ) N X + ˆp Y ( ˆp Y ) N Y = The Z-statistic is ˆσ = The critical value under 5% significant level is.96. Thus we can reject the null of no difference. 0. Do you wake up more readily on Mondays than on Fridays? To test this hypothesis, you collect data about how many minutes you let your radio alarm play before finally shutting it off and getting up. You record this data for 20 consecutive Mondays and Fridays. Under the null hypothesis that there is no difference in wake-up times, shuffling the labels on the data to scramble the Monday data in with the Friday data should not change the average difference in wake up times between the two groups of observations. The images on the following page show the results of shuffling the data 50,000 times and recording the resulting difference D between the two sets of observations ( Monday times, minus Friday times). What is the p-value of the data if the alternative hypothesis is that the Monday times are lower than the Friday times? The p value is 0.04, which we obtain by reading off the CDF plot zoomed in, from panel three of Figure. 8

9 Histogram of d CDF plot Density P(D<d) d d CDF plot (zoomed in) P(D<d) d Figure : The observed difference in wake-up times between Mondays and Fridays is shown as a vertical dashed line. 9

10 Critical values If Z N(0, ) is a standard normal random variable, the following statements hold: P (Z >.28) = 0% P (Z >.64) = 5% P (Z >.96) = 2.5% P (Z > 2.33) = % P (Z > 2.58) = 0.5% z 0

Lecture 23. STAT 225 Introduction to Probability Models April 4, Whitney Huang Purdue University. Normal approximation to Binomial

Lecture 23. STAT 225 Introduction to Probability Models April 4, Whitney Huang Purdue University. Normal approximation to Binomial Lecture 23 STAT 225 Introduction to Probability Models April 4, 2014 approximation Whitney Huang Purdue University 23.1 Agenda 1 approximation 2 approximation 23.2 Characteristics of the random variable: