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PSY 216 Assignment 7 Answers 1. Problem 2 from the text Describe the distribution of sample means (shape, expected value, and standard error) for samples of n = 36 selected from a population with a mean of μ = 100 and a standard deviation of σ = 12. According to the central limit theorem, the distribution should be normally distributed with a mean = 100 and a standard deviation = 12 / √36 = 2 2. Problem 6 from the text For a population with a mean of μ = 70 and a standard deviation of σ =210, how much error, on average, would you expect between the sample mean (M) and the population mean for: a. a sample of n = 4 scores: σM = σ / √n = 20 / √4 = 10 b. a sample of n = 16 scores: σM = σ / √n = 20 / √16 = 5 c. a sample of n = 25 scores: σM = σ / √n = 20 / √25 = 4 3. Problem 8 from the text If the population standard deviation is σ = 8, how large a sample is necessary to have a standard error that is: a. less than 4 points? σM = σ / √n √n = σ / σM n = (σ / σM)2 n = (8 / 4)2 = 4 b. less than 2 points? n = (σ / σM)2 n = (8 / 2)2 = 16 c. less than 1 point? n = (σ / σM)2 n = (8 / 1)2 = 64 4. Problem 13 from the text A random sample is obtained from a normal population with a mean of μ = 30 and a standard deviation of σ = 8. The sample mean is 33. a. Is this a fairly typical sample mean or an extreme value of a sample of n = 4 scores? The distribution of sample means will be normally distributed because the population is normally distributed. The distribution of sample means has a mean of 30 and a standard deviation of 8 / √4 = 4. Convert the raw score, 33, to a z score = (33 – 30) / 4 = 3/4 = 0.75. The central 95% of the unit normal distribution is between z = ±1.96. Since this z-score (0.75) is within the central 95% of the distribution, it is neither extreme or unusual. b. Is this a fairly typical sample mean or an extreme value for a sample of n = 64 scores? The distribution of sample means is normally distributed because the population is normally distributed; it is also likely to be normally distributed by the central limit theorem. It has a mean of 30 and a standard deviation of 8 / √64 = 1. Convert the raw score, 33, to a z score = (33 – 30) / 1 = 3/1 = 3. The central 95% of the unit normal distribution is between z = ±1.96. Since this z-score (3) is outside the central 95% of the distribution, it is extreme and unusual. 11. Problem 14 from the text The population of IQ scores forms a normal distribution with a mean of μ = 100 and a standard deviation of σ = 15. What is the probability of obtaining a sample mean greater than M = 97, a. for a random sample of n = 9 people? σM = σ / √n = 15 / √9 =5 z = (X – μ) / σM = (97 – 100) / 5 = -0.6 In a table of areas under the unit normal distribution, find the area above that corresponds to z = -0.60. If the table does not contain negative z-scores, flip the sign and side – find the area below that corresponds to z = 0.60. That area is 0.7257. b. for a random sample of n = 25 people? σM = σ / √n = 15 / √25 =3 z = (X – μ) / σM = (97 – 100) / 3 = -1.0 In a table of areas under the unit normal distribution, find the area above that corresponds to z = -1.00. If the table does not contain negative z-scores, flip the sign and side – find the area below that corresponds to z = 1.00. That area is 0.8413. 12. Problem 20 from the text The average age for licensed drivers in the county is μ = 40.3 years with a standard deviation of σ = 13.2 years. a. A researcher obtained a random sample of n = 16 parking tickets and computed an average age of M = 38.9 years for the drivers. Compute the z-score for the sample mean and find the probability of obtaining an average age this young or younger for a random sample of licensed drivers. Is it reasonable to conclude that this set of n = 16 people is a representative sample of licensed drivers? If the distribution of sample means is normally distributed, it has a mean of 40.3 years and a standard deviation of 13.2 / √16 = 3.3 years. Convert the raw score, 38.9, to a z score = (38.9 – 40.3) / 3.3 = -1.4/3.3 = -0.42. The area below a z-score = -0.42 is the same as the area above a z-score of 0.42. The area above z = 0.42 is .3372. Since the z-score is within ±1.96, it is reasonable to conclude the this sample is representative of licensed drivers. b. The same researcher obtained a random sample of n = 36 speeding tickets and computed an average age of M = 36.2. Compute the z-score for the sample mean and find the probability of obtaining an average age this young or younger for a random sample of licensed drivers. Is this a reasonable to conclude that this set of n = 36 people is a representative sample of licensed drivers? The distribution of sample means is likely normally distributed due to the central limit theory. It has a mean of 40.3 and a standard deviation of 13.2 / √36 = 2.2. Convert the raw score, 36.2, to a z score = (36.2 – 40.3) / 2.2 = -4.1/2.2 = -1.864. The area below a z-score = -1.864 is the same as the area above a z-score of 1.864. The area above z = 1.864 is .0314.The central 95% of the unit normal distribution is between z = ±1.96. Since this z-score (-1.864) is within the central 95% of the distribution, it is reasonable to conclude the this sample is representative of licensed drivers.