TutWeek9_F23

pdf

School

Wilfrid Laurier University *

*We aren’t endorsed by this school

Course

285

Subject

Statistics

Date

Jan 9, 2024

Type

pdf

Pages

Uploaded by CaptainTurkey2905

EC285 A,B Tutorial Questions Week9 Sampling Distributions Edda Claus EC285 A,B - Fall 2023 17 November 2023 1. Of all cars on the highway, 80% exceed the speed limit. What proportions of speeders might we see among the next 50 cars? Answer We want to find the distribution of the proportion of the next 50 cars that may be speeding on the highway. 80% of all cars on the highway are speeding. 10% Condition: The 50 cars can be considered a representative sample of all cars on the highway, and 50 is less than 10% of all cars on the highway. Success/Failure Condition: np = 50(0 . 80) = 40, nq = 50(0 . 20) = 10 are both at least 10. Therefore, the sampling model for the proportion of speeders in 50 cars has mean 0.80 and standard deviation of SD (ˆ p ) = p pq n = q 0 . 8 · 0 . 2 50 ≈ 0 . 057 The model for ˆ p is N (0 . 8 , 0 . 057. According to the Normal model, we expect 68% of the samples of 50 cars to have pro- portions of speeders between 0.743 and 0.857, 95% of the samples to have proportions between 0.686 and 0.914, and 99.7% of the samples to have proportions between 0.629 and 0.971. 2. Suppose that 52% of voters plan to vote “Yes” on a referendum on increasing school funding. We poll a random sample of 300 voters. What might the percentage of yes- voters appear to be in our poll? 1

Answer We want to find the distribution of the proportion of 300 voters that may vote “Yes” on the referendum. 52% of voters plan to vote “Yes”. 10% Condition: The 300 voters can be considered a representative sample of all voters, and 300 is less than 10% of all voters. Success/Failure Condition: np = 300(0 . 52) = 156, nq = 300(0 . 48) = 144 are at least 10. Therefore, the sampling model for the proportion of “Yes”-voters in a sample of 300 has mean 0 . 52 and standard deviation of SD (ˆ p ) = p pq n = q 0 . 52 · 0 . 48 300 ≈ 0 . 029 The model for ˆ p is N(0.52, 0.029). According to the Normal model, we expect 68% of the samples of 300 voters to have proportions of “Yes” voters between 0 . 491 and 0 . 549, 95% of the samples to have proportions between 0 . 462 and 0 . 578, and 99 . 7% of the samples to have proportions between 0 . 433 and 0 . 607. 3. A new sales gimmick has 30% of the M&M’s covered with speckles. These “groovy” candies are mixed randomly with the normal candies as they are put into the bags for distribution and sale. “Groovy” M&M’s are supposed to make up 30% of the candies sold. In a large bag of 250 M&M’s, what is the probability that we get at least 25% “groovy” candies? Answer We want to find the probability of getting a bag of 250 M&M’s that has at least 25% “groovy” candies. 10% Condition: The 250 M&M’s can be considered a representative sample of all M&M’s, and 250 is less than 10% of all M&M’s. Success/Failure Condition: np = 250(0 . 30) = 75, nq = 250(0 . 70) = 175 are at least 10. Therefore, the sampling model for the proportion of “groovy” M&M’s in a sample of 250 has mean 0 . 30 and standard deviation of SD (ˆ p ) = p pq n = q 0 . 3 · 0 . 7 250 ≈ 0 . 029 The model for ˆ p is N(0.3, 0.029). According to the Normal model, the probability that our bag contains at least 25% “groovy” M&M’s is approximately 95.8%. 4. At birth, babies average 7.8 pounds, with a standard deviation of 2.1 pounds. A random sample of 34 babies born to mothers living near a large factory that may be polluting the air and water shows a mean birth weight of only 7.2 pounds. Is that unusually low? 2

Answer We are interested in the probability that a sample of babies has mean birth weight below 7.2 pounds. At birth, babies average 7.8 pounds, with a standard deviation of 2.1 pounds. The model for birth weights should be roughly unimodal and symmetric, if not Normal. Random Sampling Condition: The 34 babies were sampled randomly. Independence Assumption: It’s reasonable to think that the weights of the randomly sampled babies will be mutually independent. 10% Condition: As long as more than 340 babies were born to mothers living in the vicinity of the factory, 34 babies represent less than 10% of all babies. As the model for all babies is unimodal and symmetric, the Central Limit Theorem applies, especially because the sample size, 34 babies, is comparatively large. Under these conditions, the sampling distribution of is Normal with mean pounds and standard deviation SD ( y ) = σ √ n = 2 . 1 √ 34 ≈ 0 . 36 pounds. According to the Normal model, only about 4.75% of samples of 34 babies are expected to have mean birth weights below 7.2 pounds. The samples of babies near the factory appear to have a usually low mean birth weight. 5. Assume that the true proportion of college students who have taken at least one online course is 0.22. Suppose a sample of 500 college students is selected. a. What would you expect the shape of the sampling distribution for the sample pro- portion to look like? b. What would be the mean of this sampling distribution? c. What would be the standard deviation of this sampling distribution? d. If the sample size is increased to 1000, how would the sampling distribution change? Answer a. Normal; b. 0.22; c. The mean would not change, and the standard deviation would decrease to 0.0131. 6. Assume that the true proportion of college students who do not have dental insurance is .85. Suppose that a sample of 200 college students is selected. a. Is the success/failure condition satisfied? Explain. 3

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

b. What is the probability that the sample proportion will be at least 0.85? c. What is the probability that the sample proportion will be less than 0.80? d. What is the probability that the sample proportion will be between 0.80 and 0.90? Answer a. Yes, the success/failure condition is satisfied: np = 200 · 0 . 85 = 170 > 10 nq = 200 · 0 . 15 = 30 > 10; b. 0.5; c. 0.0239; d. 0.9522; 7. The net worth of families in New Brunswick in thousand dollars has a mean of 220 with a standard deviation of 40. A sample of families in New Brunswick is interviewed and the sample mean net worth is computed. a. Assuming a large enough sample size, what would you expect the shape of the sampling distribution for the sample mean to look like? b. What would be the mean of this sampling distribution? c. If the sample size is 50, what would be the standard deviation of this sampling distribution? d. If the sample size is 100, what would be the standard deviation of this sampling distribution? Answer a. Normal; b. 220; c. 5,66; d. 4. 8. Assume that new eco-friendly plastic bags have a mean tensile strength of 90 MPa with a standard deviation of 10 MPa. Suppose that a sample of 64 bags is selected. a. What is the probability that the sample mean will be at least 90 MPa? 4

b. What is the probability that the sample mean will be between 87.5 MPA and 92.5 MPa? c. What is the probability that the sample mean will be less than 91MPa? d. What is the probability that the sample mean will be more than 93.75 MPa? Answer a. 0.5; b. 0.9544; c. 0.7881; d. 0.0013. 9. The number of hours that the WLU students spend on social networking sites per day is normally distributed with a mean of 4 hours and a variance of 1.44. a. What is the probability that a randomly selected WLU student spend more than 2 hours but less than 5.5 hours per day on social networking sites? b. Suppose that Hong spends 2 hours and 15 minutes per day on social networking sites. At which percentile is she in the distribution for the WLU students? c. Suppose you are going to take a random sample of five WLU students. What is the probability that none of them in the sample spends more than 3.8 hours per day on social networking sites? Answer a. T ∼ (mean = 4 , sd = 1 . 2 z − score (2) = 2 − 4 1 . 2 = − 1 . 67 p ( − 1 . 67 < Z ) = 0 . 0475 z − score (5 . 5) = 5 . 5 − 4 1 . 2 = 1 . 25 p ( Z < 1 . 25) = 0 . 8944 P (4 < X < 5) = P ( − 1 . 67 < Z < 1 . 25) = 0 . 8944 − 0 . 0475 = 0 . 8469 b. Convert to z-score: z = 2 . 25 − 4 1 . 2 = − 1 . 46 P ( Z < − 1 . 46) = 0 . 0721 So she is at the 7 . 2 nd percentile c. This is a binomial. Sampling 5 students, none spend more than 3 . 8 hours per day. If none spend more than 3.8 hours a day, all must spend less. The probability of 5

that is: p ( less ) ∗ p ( less ) ∗ p ( less ) ∗ p ( less ) ∗ p ( less ) = p ( less ) 5 P ( X < 3 . 8) = P ( Z < (3 . 8 − 4) 1 . 2 ) = P ( Z < 0 . 16667) = 0 . 5675 P (none more than 3 . 8) = P (all less than 3 . 8) = 0 . 5675 5 = 0 . 05886. 10. Suppose you wanted to create an alternative to the 64-95-99.7 Rule for the normal distribution where the percentages (rather than the standard deviations) came out to nice round numbers. Fill in the missing values. a. About 25% of observations are within ± 0 . 32 s.d. of the mean; b. About 50% of observations are within ± 0 . 675 s.d. of the mean; c. About 75% of observations are within ± 1 . 15 s.d. of the mean; 11. You have two random variables: X 1 ∼ N ( µ 1 , σ 1 ) and X 1 ∼ N ( µ 2 , σ 2 ). X 1 and X 2 are independent. Define a new random variable W = aX 1 + bX 2 . We learned that a linear combination of two normal variables also follows a normal distribution. Write down the distribution of W , in terms of the parameters of X 1 and X 2 . ( i.e. , W ∼ (? , ?) Answer W = aX 1 + bX 2 E ( W ) = E ( aX 1 + bX 2 ) = aE ( X 1 ) + bE ( X 2 ) = aµ 1 + bµ 2 V ar ( W ) = V ar ( aX 1 + bX 2 ) = a 2 V ar ( X 1 ) + b 2 V ar ( X 2 ) + 2 · a · b · Cov ( X 1 , X 2 ) = a 2 σ 1 + b 2 σ 2 + 0 W ∼ N ( aµ 1 + bµ 2 , √ a 2 σ 1 + b 2 σ 2 ) 12. Given the following, and assuming that X and Y are independent, find the distribution of W: X ∼ N ( µ X , σ 2 X ) Y ∼ N ( µ Y , σ 2 Y ) W = a + bX + cY Answer E ( W ) = E ( a + bX + cY ) = a + bE ( X ) + cE ( Y ) = a + bµ X + cµ Y V ar ( W ) = V ar ( a + bX + cY ) = V ar ( bX + cY ) = b 2 V ar ( X )+ c 2 V ar ( Y )+2 bcCov ( X, Y ) = b 2 σ 2 X + c 2 σ 2 Y 13. Given the information in the graphs about two normally distributed independent ran- dom variables X and Y , draw the distribution of W if W = ( X – Y ). Include the parameters of the distribution of W in your answer. 6

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

Answer From reading the graphs, I would say : X ∼ N ( − 10 , 2); Y ∼ N (10 , 2) E ( X ) = − 10; V ar ( X ) = 2 2 = 4 E ( X ) = 10; V ar ( X ) = 2 2 = 4 W = X − Y W ∼ N (sum of two normals) E ( W ) = E ( X − Y ) = E ( X )– E ( Y ) = − 10 − 10 = − 20 V ar ( W ) = V ( X − Y ) = 1 2 · V ar ( X ) + ( − 1) 2 · V ar ( Y ) + 2 · 1 · ( − 1) COV ( X, Y ) = 1 · 4 + 1 · 4 + 0 = 8 So, SD ( W ) = √ 8 = 2 . 82 So, W ∼ N ( − 20 , 2 . 82) 7

TutWeek9_F23

Related Documents