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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.