Download 10.4 Power Type Errors Example - Mrs. McDonald

1. A research psychologist plans to administer a test designed to measure self-confidence to a random sample of fifty
professional athletes. The psychologist theorizes that professional athletes tend to be more self-confident than
others. Since the national norm of the test is known to be 72, the theory may be partially validated if it can be shown
that the mean score for all professional athletes, μ, exceeds 72. Suppose the sample mean and standard deviation of
the fifty scores are
x = 74.1 and s = 13.3.
a) Indicate the null and alternative hypothesis.
b) Calculate the z-score and probability for the sample mean. Do these data support the research hypothesis of the
psychologist? Use α = .10.
c) On the normal curve, neatly label μ, x , and the three standard deviations in each direction.
d) What is the type one error in this case?
e) Find P(Type I error).
f)
Find x that would separate the rejection region from the rest of the region using α = .10. Should you use -1.28 or
1.28?
g) On the above normal curve, shade and label the rejection region.
h) What is the Type II error in this case?
i)
Suppose that the actual mean is 75. Redraw and label the normal curve with μ = 75 and the new standard
deviations. Label the old mean and shade the old rejection region.
j)
Calculate the probability β of a Type II error for the test in #1 if the mean score for professional athletes, μ,
actually equals 75. This is the probability of failing to conclude that the professional athletes’ mean score
exceeds the national norm when, in fact, their mean score is as large as 75.
2. A machine is set to produce bolts with a mean length of 1 inch. Bolts that are too long or too short do not meet the
customer’s specifications and must be rejected. To avoid producing too many rejects, the bolts produced by the
machine are sampled from time to time and tested to see whether the machine is still operating properly, i.e.,
producing bolts with a mean length of 1 inch. Suppose fifty bolts have been samples and
= .04 inch.
x = 1.02 inches long and σ
a) Indicate the null and alternative hypothesis.
b) Calculate the z-score and probability for the sample mean. Do these data support the alternative hypothesis?
Use α = .01.
c) On the normal curve, neatly label μ, x , and the three standard deviations in each direction.
d) What is the type one error in this case?
e) Find P(Type I error).
f)
Find x that would separate the rejection region from the rest of the region using α = .01. Should you use -2.575
or 2.575?
g) On the above normal curve, shade and label the rejection region.
h) What is the Type II error in this case?
i)
Suppose that the actual mean is 1.005 inches long. Redraw and label the normal curve with μ = 1.005 and the
new standard deviations. Label the old mean and shade the old rejection region.
j)
Calculate the approximate probability β of a Type II error for the test in #2 if the mean score, μ, actually equals
1.005 inches.
Answers:
H0: μ = 72
Ha: μ > 72
[look up α = .10]
Rejection region: z > 1.28
Test Statistic:
z
x  72

n

74.1  72
 1.12
13.3
50
Since 1.12 does not fall within the rejection region for α = .10, we don’t support the psychologists theory.
The rejection region for the test in #1 is z > 1.28 or, equivalently, values of
Above the H0: μ = 72. This point is shown on the
x that lie mor e than 1.28sample st dev.