STT 315, Section 201 Worksheet 4 07/25/2014 1. Let X be the following random variable X 1 P(X) 0.25 a) Calculate E(X) and S.D(X) E(X)=2.5 S.D.(X)=1.118 2 0.25 3 0.25 4 0.25 b) If we take a sample of size n=2, what are all the possible combinations of samples with replacement we might have? (1,1) , (2,1), (3,1) , (4,1), (1,2), (2,2), (3,2), (4,2), (1,3), (2,3), (3,3), (4,3), (1,4), (2,4), (3,4), (4,4) c) Calculate the means for each of the samples above and what is the probability of getting a particular sample? Are the samples equally likely? Means : 1 1.5 2 2.5 1.5 2 2.5 3 2 2.5 3 3.5 2.5 3 3.5 4 The probability of each of the samples is 1/16, so they are equally likely. d) What is the probability distribution for the sample mean? X P(X ) 1 1/16 1.5 2/16 2 3/16 2.5 4/16 3 3/16 e) Draw a histogram for the distribution of the means. 3.5 2/16 4 1/16 STT 315, Section 201 Worksheet 4 07/25/2014 2. Let X be the following random variable X 2 4 10 20 P(X) 0.6 0.2 0.1 0.1 a) Calculate E(X) and S.D(X) E(X)=5 SD(X)=5.532 b) If we take a sample of size n=2, what are all the possible combinations of samples with replacement we might have? (2,2), (4,2), (10,2), (20,2), (2,4), (4,4), (10,4), (20,4), (2,10), (4,10), (10,10), (20,10), (2,20), (4,20), (10,20), (20,20) c) Calculate the means for each of the samples above and what is the probability of getting a particular sample? Are the samples equally likely? P(sample) x̅ P(sample) x̅ P(sample) x̅ P(sample) x̅ 2 0.6*0.6=0.36 3 0.6*0.2=0.12 6 0.1*0.6=0.06 11 0.1*0.6=0.06 3 0.6*0.2=0.12 4 0.2*0.2=0.04 7 0.1*0.2=0.02 12 0.1*0.2=0.02 6 0.6*0.1=0.06 7 0.2*0.1=0.02 10 0.1*0.1=0.01 15 0.1*0.1=0.01 11 0.6*0.1=0.06 12 0.2*0.1=0.02 15 0.1*0.1=0.01 20 0.1*0.1=0.01 d) What is the probability distribution for the sample mean? X P(X ) 2 0.36 3 0.24 4 0.04 6 0.12 7 0.04 10 0.01 11 0.12 d) Draw a histogram for the distribution of the means. 12 0.04 15 0.02 20 0.01 STT 315, Section 201 Worksheet 4 07/25/2014 Confidence Intervals: 1. What is the confidence level for each of the following intervals? a) x̅ ± 1.96σ/√n normalcdf(-1.96,1.96)~95% b) x̅ ± 1.645σ/√n normalcdf(-1.645,1.645)~90% c) x̅ ± 2.575σ/√n normalcdf(-2.575,2.575)~99% d) x̅ ± 0.99σ/√n normalcdf(-0.99,0.99)~68% 2. A random sample of n measurements was selected from a population with unknown mean μ and std deviation σ. Calculate a 95% confidence interval for each of the following: a) n=80, x̅ =32, s2=14 (32-1.96*√ 14/80, 32+1.96*√ 14/80)= (31.18,32.82) b) n=210, x̅ =107, s2=23 (107-1.96*√(23/210),107+1.96*√(23/210))=(106.35,107.65) c) n=105, x̅ =13, s=0.3 (13-1.96*0.3/√(105),13+1.96*0.3/√(105))= (12.943,13.057) 3. A random sample of 92 observations produced a mean of 25.5 and a std. deviation s=2.8. Find a 90% confidence interval for μ. (25.5-1.645*2.8/√(92),25.5+1.645*2.8/√(92))= (25.02,25.98) 4. Each child of 61 low income children was administered a language and communication exam. The sentence complexity scores had a mean of 7.62 and a standard deviation of 8.94. a) From the sample, estimate the true mean sentence complexity score of all low-income children. μ̂ = x̅ = 7.62 b) Form a 90% confidence interval of μ. (7.62-1.645*8.94/√(61), 7.62+1.645*8.94/√(61)) = (5.7372,9.5028) c) Suppose the true mean sentence complexity score of middle-income children is known to be 15.55. Is there evidence that the true mean for low-children differs from 15.55? Yes, because 15.55 isn’t in the interval in (b) 5. Let t0 be specific value of t. Use the table of t-critical values of t below to find t0 values such that the following statements are true. t0=2.160 a) P(-t0<t<t0)=0.95, df=13 b) P(t≤ t0)=0.01, df=20 t0=-2.528 t0=1.782 c) P(t≤ -t0 or t≥ t0)=0.10 , df=12 d) P(t≤ -t0 or t≥ t0)=0.01, df=10 t0=3.169 6. The random sample shown below was selected from a normal distribution. 4, 8, 7, 10, 3, 4 a) Construct a 90% CI. T-Interval (3.7321,8.2679) b) If the sample is based on n=35, how does the CI change? Since n≥ 30, Z-Interval x̅ =6, Sx=2.757, n=6 (5.2335,6.7665) STT 315, Section 201 Worksheet 4 07/25/2014 7. An observational study of teams fishing for the red shiny lobster in a certain body of water was conducted and the results published in a science magazine. One of the variables of interest was the average distance separating traps called trap-spacing, deployed by the same team of fisherman. Trap spacing measurement (in meters) for a sample of 7 teams of fishermen. Variable of interest is the mean trap spacing. (data: 95, 97, 107, 96, 81, 72, 84) a) What is the target parameter? The target parameter is the mean (μ) distance separating traps. b) Compute a point estimate for the population parameter. μ̂ = Mean of data= 90.2857 c) Why can’t we use a z-statistic? Since n<30. d) Compute 95% CI. T-Interval (79.347, 101.22) e) What are the conditions that must be satisfied? That the population form which the data is obtained should follow a normal distribution. 8. In 2010, a survey of 400 regions, had 200 over-estimated marketed values. Suppose you want to estimate p, the true proportion of homes that have over-estimated marketed values, a) Find p̂ , the point estimate of p. p̂ =0.5 b) Describe the sampling distribution of p̂ . p̂ ~ N( 0.5, √ (0.25/400)) i.e. p̂ ~ Normal (0.5, 0.025) c) Find a 90% CI for p (0.45888,0.54112) e) Suppose a researcher claims that p=0.44 Is the claim believable? No, Since 0.44 doesn’t belong to the above interval. 9. The paralyzed Veterans of America is a philanthropic organization that relies on contributions. They send free mailing labels and greeting cards to potential donors on their list and ask for voluntary contribution. To test a new campaign they recently sent letters to a random sample of 100,000 potential donors and received 4781 donations. a) Give a 95% confidence interval for the true proportion of those from their entire mailing list who may donate. 1-PropZInterval : (0.04649,0.04913) b) A staff member thinks that the true rate is 5%. Given the confidence interval you found, do you find that percentage plausible? No, 5% doesn’t belong in the above interval 10. In a poll taken in March of 2007, Gallup asked 1006 national adults whether they were baseball fans. 36% said they were. A year previously 37% of a smaller size sample had reported being baseball fans. STT 315, Section 201 Worksheet 4 07/25/2014 a) Find the margin of error for the 2007 poll if we want 90% confidence in our estimate of the percent of national adults who are baseball fans. ME=1.645*√ (0.36*0.64/1006)=0.025 b) Explain what the margin of error means. It is the error region about the point estimate. c) If we wanted to be 99% confident, would the margin of error be larger or smaller? Larger d) Find the margin of error for 99% confidence level. ME= 2.576*√ (036*0.64/1006)=0.039 e) In general, all other aspects of the situation remain the same; will smaller margins of error produce greater or less confidence in the interval? less f) Do you think there’s been a change from 2006 to 2007 in the real proportion of national adults who are baseball fans? No, since it belongs to (0.321,0.399) 11. Several factors are involved in the creation of a confidence interval. Among them are the sample size, the level of confidence, and the margin of error. Which statements are true? a) b) c) d) e) f) For a given sample size, higher confidence means a smaller margin of error.(F) For a specified confidence level, larger samples provide smaller margins of error.(T) For a fixed margin of error, larger samples provide a greater confidence.(T) For a given confidence level, halving the margin of error requires a sample twice as large.(F) For a given sample size reducing the margin of error will mean lower confidence.(T) For a certain confidence level, you can get a smaller margin of error by selecting a bigger sample.(T) g) For a fixed margin of error, smaller samples will mean lower confidence.(T) h) For a given confidence level, a sample 9 times as large will make a margin of error one third as big.(T)

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