Fall 2009 Stats 250 Exam 1 with detailed Solutions and Explanations
Fall 2009 Stats 250 Exam 1 with detailed Solutions and Explanations STATS 250
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Fall 2009 Stats 350 Exam 1 Explanations 1. a. The population of interest for this survey is: All higher education professionals in the United States. The second and fourth options are incorrect because they refer to the specific sample. The third option is incorrect because higher education professionals not institutions were surveyed. b. The value 0.693 corresponds to: , statistic, and sample proportion. Circle all and only these 3. 2. This is an example of confounding. The proposed relationship between coffee and heart attacks may be due entirely or in part to the tendency for smokers to both have more heart attacks and drink more coffee. 3. No, this is a time-plot not a histogram. 4. a. Length of dinner in minutes is a quantitative continuous variable. Number of evenings eating dinner together is a quantitative discrete variable. Eating dinner together or not last Sunday is a categorical variable. The third question is categorical because it divides response into groups ‘yes’ and ‘no’. Since they can be measured, the first two bullets are both quantitative because they can be measured. The number of evenings spent eating dinner as a family last week can take only a limited number of values (0, 1, 2, … 7) so it is discrete. In contrast, since fractions and decimals make sense for average time spent eating dinner this first choice is continuous. b. i. The longest reported time to prepare a home dinner for the family was about 65 minutes, while 25% of those responding reportedly take at least 40 minutes to prepare dinner. The longest reported time is the outlier represented as a dot in the boxplot while “25% at least” indicates the third quartile (top of the box). ii. The average distance between the times to prepare a home dinner is roughly 8.9 minutes. Incorrect, standard deviation is not the average distance between observed times. The times to prepare a home dinner vary by about 8.9 minutes, on average. Incorrect, again this one is missing the key idea of varying from the mean. On average, the times to prepare a home dinner varied from the mean by about 8.9 minutes. Correct. 13 5. Stay-at-home Mom Working Mom Total Rate herself at the top 212 198 410 Rate herself below the top 188 402 590 Total 400 600 1000 a. One option is to fill in the missing totals first: 1000 – 600 = 400, and 1000 – 410 = 590. Then find the other missing value as either 400 – 212 = 188 or 590 – 402 = 188. b. Working Status plays the role of the explanatory variable. c. ‘Randomly selected’ means use the entire table: P(Rate herself at the top) = 410/1000 = 0.41. d. P(Rate herself at top | Working Mom) = 198/600 = 0.33. Use only the ‘Working Mom’ column since this information is given as known. e. Yes, there appears to be an association, because … ‘working status’ does not appear to be independent from how a mother rates herself. We can see this from P(Rate herself at the top | Working Mom) ≠ P(Rater herself at the top), 0.33 ≠ 0.41. … or … We can see this from… P(Rate herself at the top | Working Mom) ≠ P(Rater herself at the top | Stay-at-home mom) 0.33 ≠ 0.53 = 212/400. 6. a. This question is asking us to describe the sampling distribution of the sample proportion ̂. Here ( ) n = 400 and p = 0.45, ̂ is approximately( √ ) ( ) b. We must also verify that the sample size is large enough for the normal distribution to be a good approximation to the binomial. This is true since the conditions below are satisfied: np = 400*0.45 = 180 > 10 and n(1-p) = 400(0.55) = 220 > 10. c. Calculate the z-score for ̂ : (0.5 – 0.45)/0.0249 = 2.01. So, P( > 2.01) = 0.0222 = 2.22%. Want more review? Visit www.rossmanchance.com/applets/NormalCalcs/NormalCalculations.html 14 7. This is a binomial distribution with n = 7 trials and probability of ‘success’ (student coming away a ‘winner’) p = 0.20. The number of students who come away a winner, X, is Binomial(7, 0.2). We want to find P(X > 2), but we can use the complement rule to make this simpler: P(X > 2) = 1 – P(X < 2) = 1 – P(X = 1) – P(X = 0). Now, use the binomial probability formula (page 1 of formula card) to compute: P(X = 1) =( ) ( ) ( ) and P(X = 0) =( ) ( ) ( ) Therefore, P(X > 2) = 1 – 0.3670 – 0.2097 = 0.4233. Want more review of binomial probability problems? Visit www.ltcconline.net/greenl/java/Statistics/Binomial/Binomial.htm (in Internet Explorer or Chrome) 8. a. The confidence interval for p is (0.32, 0.48) which we know is found as ̂ margin of error. The midpoint of the interval is (0.32 + 0.48)/2 = 0.40, so ̂ = 0.40 and the margin of error is 0.08 (find this as 0.48 – 0.40 = 0.08 or 0.40 – 0.32 = 0.08). b. From above, we have ̂ = 0.40 = #prefer tea/250. Solve for ‘#prefer tea’ (call this x if it helps) to get, #prefer tea = 0.40 * 250 = 100. c. Only the first choice is correct: If the sampling procedure were repeated many times, then approximately 99% of the resulting intervals would contain the population proportion of all customers that prefer tea over coffee. One reason each of the other two is incorrect is that both refer to this specific interval while the confidence level always refers to the procedure. d. The desired margin of error is m = 4% = 0.04 and the multiplier for a 99% confidence level is z* = 2.58 (look up 0.995 in table A.1). Now use the sample size formula from page 2 of the formula card: ( ) ( ) ( ) Always round a sample size up to the nearest whole number, so a sample of size 1041 people is needed. 9. a. ● This proves that the politician’s claim is incorrect. Not reasonable, as it is too strong. ● The politician’s claim could be correct, but the sample size was too small to detect it. Reasonable, with a larger sample size we would have more power to detect the result. ● The null hypothesis should be rejected. Not reasonable, since 0.11 > 0.10. b. The sample size is too small for the normal approximation: n(1 – p )0= 20(1 -0.7) = 6 < 10. In this case, the p-value is computed using the small sample binomial test – the distribution is used to compute the p-value is Binomial(n = 20, p = 0.7). 15 10. a. µ = E(X) = 0(0.4) + 1(0.5) + 2(0.1) = 0.7 visits. Either µ or E(X) is fine for the symbol. No need to round an expected value. b. ‘At least 1’ means one or more visits: P(X > 1) = P(X = 1) + P(X = 2) = 0.5 + 0.1 = 0.6. c. P(X = 2 | X > 1) = P( X = 2 & X > 1) / P(X > 1) = P(X = 2)/P(X > 1) = 0.1/0.6 = 1/6 = .1667. Note that P( X = 2 & X > 1) = P(X = 2) since 2 > 1. 11. a. H 0 p = 0.25 H a p > 0.25 b. See the image to the right. To find the p-value: P(Z > 1.55) = 1 – P(Z < 1.55) = 1 - .9394 = 0.0606 c. False. The normal distribution is being used as an approximation to the binomial distribution. The categorical variable “disturbed sleep due to money concerns” (yes/no) cannot be normally distributed. 12. a. Fail to Reject H s0nce 0.08852 > 0.05. b. Reject H , since a Type I error is incorrectly rejecting H . 0 0 c. Reject H 0his is what statistically significant means. 13. a. Find Jim’s z-score: b. ‘At most 170’ means 170 or less. Find the z-score for 170: . Now, P( X < 170) = P (Z < 0.5714) = 0.7161 or about 71.61%. The picture below left will help. c. Again, a picture will be helpful here, see above right. Find the z-score giving the value closest to 60% = 0.60 inside table A.1: z = 0.25 corresponds to 0.5987 (which is closer than 0.6026). Now, use the equation ( ) ( ) pounds. Want more review? Visit www.rossmanchance.com/applets/NormalCalcs/NormalCalculations.html 16 Fall 2009 Exam 1 without answers 1. UseTwitter? Despite the hype surrounding Twitter, most higher education professionals are not using the microblogging service at all. In a survey of 1,958 highe r education professionals conducted by Faculty Focus in July and August 2009, 69.3 percent of respondents said they do not use Twitter in any capacity. a. The population of interest for this survey is:  Circle one. • All higher education professionals in the United States. • The 1,958 higher education professionals who answered the question. • All higher education institutions in the United States. • The 69.3% who answered "Yes" to the question b. Which of the following pertain to the value of 0.693 or 69.3%? Circle all that apply.  Populatipnroportion parameter statistic sample proportion 2. Coffee and Heart Attacks: A recent medical study found that people who drink more than 4 cups of coffee a day have more heart attacks than people who drink less coffee or no coffee. This led some doctors to suspect that coffee may be a contributing factor in causing heart attacks. However, more careful analysis of the data showed that heavy coffee drinkers tend to smoke more than other people. This is an example of:  Circle your answer: the placebo effect a response bias confounding the double-blind technique 3. Applicationlevels: The graph at the right was part of the St. Olaf Annual Report, in the section providing information about applications to the college. Based on this graph, is it appropriate to say the distribution of application levels is skewed to the left?  Circleone: Yes No Briefly explain: 135 4. EatingHabits: A study was conducted to learn about eating habits for American families. A random sample of 200 families was selected and an adult head of household was asked to complete a survey. a. Listed below are some questions that were asked as part of the survey. For each, determine whether it is categorical, quantitative discrete, or quantitative continuous. [1 pt each] Clearly circle your answer. • How long would you say dinner usually lasts when your family eats dinner together (in minutes)? categoridciaslcrequantitative te quac ntittnveous • In the last week, how many evenings did your family eat dinner together? categoridciaslcrequantitative te quac ntittnveous • Did your family eat dinner together last Sunday evening? categoridciaslcrequantitative te quac ntittnveous b. Another question on the survey asked for the time it typically takes to prepare a home dinner for the family (in minutes). Below and at the right is a summary of the responses. Use these results to answer the following questions. Mean Std. Deviation Prep_Time (minutes) 35.6 8.9 i. Complete the sentence:  The longest reported time to prepare a home dinner for the family was about ___________ minutes, while 25% of those responding reportedly take at least __________ minutes to prepare a home dinner for the family. ii. The standard deviation for preparation time is reported as 8.9. Consider the following interpretations of this value and clearly circle the correct interpretation(s).  • The average distance between the times to prepare a home dinner is roughly 8.9 minutes. • The times to prepare a home dinner vary by about 8.9 minutes, on average. • On average, the times to prepare a home dinner varied from the mean by about 8.9 minutes. 136 5. Mom Survey: A research center conducted a telephone poll based on a random sample of 1,000 women who are 18 and older and have at least one child. The women we re asked two questions: (1) whether she was a stay-at- home mom or a working mom; and (2) ho w she would rate herself as a parent on the scale of 1, 2, 3, … , 10; where “1” represents “poorest” and “10” represents “best”. The table below summarizes the poll results: the number of women who rated themselves at the top of the scale (9, 10) and below the top of the scale (1-8). Use these results to answer the following questions. Stay-at-home Mom Working Mom Total Rate herself at the top 212 19 4810 Rate herself below the top 402 Total 600 1000 a. Provide the values for the 3 missing cells to complete the above table.  b. Which variable plays the role of the explanatory variable in this study?  Circle your answer: Rating as Parent Working Status (work or stay at home) c. What is the probability that a randomly selected woman rates herself at the top as a parent?  A_iwal_:_________ d. Knowing the mom is a working mom, what is the probability she rates herself at the top as a parent?  A_iwal_:_________ e. Does there seem to be an association between whet her a mom stays-at-home or works and how she rates herself as a parent? Explain your answer with appropriate numerical support.  Yes, there does appear to be an association, Circle one: No, there does not appear to be an association, because… 137 6. VoteSupport: Consider the population of all registered Michigan voters. Suppose 45% of all registered Michigan voters support a particular candidate for political office. A random sample of n = 400 registered Michigan voters will be selected and the sample proportion that support the candidate will be computed. a. If you consider all possible random samples of 400 regi stered Michigan voters, what is the approximate distribution that describes the possible values for the samp le proportion? Be complete; give all aspects of the distribution.  b. The approximate distribution in part (a) holds only if ce rtain data conditions are met. We already have that a random sample was selected. State the remaining condition and verify that it is met for this situation.  c. What is the probability that a random sample of 400 Michig an voters will result in at least half supporting the political candidate? Show your work.  AF_wal_:________________ 7. LostWages: Suppose the probability that a college student le aves Las Vegas after 3 days as a “winner” is 0.20. Consider seven such students (strangers to one an other) who go to Las Vegas for a 3-day break. What is the probability that at least two of the seven leave Las Vegas after 3 days as a “winner”?  AF_wal_:________________ 138 8. Tea or Coffee: A national coffee chain wants to estimate the proportion of its customers that prefer tea over coffee. The chain collected a random sample of 250 customer s. All customers in the sample were asked if they preferred tea over coffee. The resulting 99% confidence interval for the population proportion of all customers that prefer tea over coffee was found to be (0.32, 0.48). a. What is the margin error for the confidence interval of (0.32, 0.48)?  AF_walr_:_________ b. How many customers in the sample reported that they preferred tea over coffee?  AF_walr_:_________ c. Which of the following are valid interpretations of the 99% confident level? (Circle all that apply.)  • If the sampling procedure were repeated many times, then approximately 99% of the resulting intervals would contain the population proportion of all customers that prefer tea over coffee. • There is about a 99% chance that the population pr oportion of all customers that prefer tea over coffee lies in the interval (0.32, 0.48). • For repeated samples of 250 customers from the sa me population, the proportion of all customers that prefer tea over coffee will fall in the interval (0.32, 0.48) 99% of the time. d. What sample size would be required to estimate the po pulation proportion with 99% confidence and a margin of error of 4%? Show your work.  AF_walr_:_________ 9. Got the vote? A politician claims that she will receive more than 70% of the vote in an upcoming election. To test this claim, a random sample of 20 voters will be surveyed. The significance level is set at 10%. a. In the sample of 20 voters, 17 stated they would vote for her. The resulting p-value is 0.11. Which of the following is a reasonable interpretation of these results?  • This proves that the politician’s claim is incorrect. • The politician’s claim could be correct, but the sample size was too small to detect it. • The null hypothesis should be rejected. b. What distribution was used to compute the p-value in part (a)?  Give all details. Distribution:____________________________________________________ 139 10. OfficeVisits: For the past 10 years, Dr. Werner has recorded the number of students who come to his Friday 12 noon to 1 PM office hours. Based on his observations he has developed the following model. X = number of students 0 1 2 Probability 0.4 0.5 0.1 a. How many students would you expect during this scheduled office hour? Include the appropriate symbol in your answer and the units.  = ________________________________ b. What is the probability of at least one student coming to office hours?  A_iwal_:_________ c. Givethat at least one student will come to office hours on a gi ven Friday, what is the probability that exactly two students will come?  A_iwal_:_________ 11. DisturbedSleep: A psychologist claims that more than 25% of a ll American adults experienced disturbed sleep in the past 6 months because they are worried about money. Let p = the population proportion of all American adults that experienced disturbed sleep in the past 6 months because they are worried about money. a. State the appropriate hypotheses to be tested in terms of p.  H 0:________________________________ H a__________________________________ b. The significance level was set at 10%. A random sample of 500 American adults resulted in 140 stating they had experienced disturbed sleep in the past 6 months because they are worried about money. The test statistic value is z = 1.55. Draw a picture that shows what the p-value corresponds to and report its value.  p-value = ________________ c. True or False? One of the conditions required for the Z test for the above hypotheses to be valid is that the model for the original population be normal. ] 2 [ True False 140 12. Decision Making: A researcher prepares a Null Hypothesis (H 0) and an Alternative Hypothesis (H a), runs an experiment, and then makes a decision to reject or fail to reject the Null Hypothesis. In each case, use the given information to determine which decision was made and circle it. a. The p-value was 0.08852 and the significance level was 0.05.  Decision: (circle one) Reject H 0 Fail to Reject 0 b. A Type I error was made.  Decision: (circle one) Reject H 0 Fail to Reject 0 c. The result was statistically significant.  Decision: (circle one) Reject H Fail to Reject H 0 0 13. Weights of Male High School Seniors: Suppose the weights for high school male seniors are normally distributed with an average of 166 pounds and a standard deviatio n of 7 pounds. Use this mode l to address the following questions. a. Jim weighs 175 pounds. How many standard deviations is he from the mean?  AF_nael_:_______________ b. What proportion of the male high school seniors weigh at most 170 pounds?  AF_nael_:_______________ c. Find the value to complete the following statement (show your work). 40% of the male high school seniors weigh _________________ pounds or more.  141
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