Download Handout 7a Example of calculating Beta

Example of Calculating Type-II error and Power of the test
In this section, you will get some practice in calculating the probability of a type-2 error for a hypothesis test.
Consider a test of H0:  = 124 versus Ha:  > 124. The test uses  = 16, =0.05 and a sample size of n = 64.
a. Describe the sampling distribution of x assuming H0 is true.
Mean (  x ) = 124 (sampling distribution mean is the same as the population mean)
Standard deviation (  x ) =
√
Shape: The sampling distribution will be Bell-Shaped (normal) with a mean of 124 and a standard deviation
of 2
Sketch the sampling distribution of x assuming H0 is true. (This is step 1 of the process of finding Type-II error)
120 122 124 126 128
b. Specify the rejection region when x is used as the test statistic. Locate the rejection region on your graph from
part a. (This is steps 2 and 3 of the process of finding Type-II error).
Step 2:
Since =0.05 and we have a one-sided test. We have a critical z-value of 1.645 (See the table of
critical values in the formula sheet notes). This means that if our observed value is more than 1.645
standard deviations above the mean, we will reject the null hypothesis. In step 2, we want to find
the corresponding value under our given normal curve (in this case, we want to find the value that
is 1.645 standard deviations above the mean of 124 where the standard deviation is 2. In the
formula sheet, I called this
.
.
So,
∗
.
In other words, if we observe a sample mean above 127.29, then we would consider that as being too
far from the hypothesised mean to have occurred by chance and that we can then conclude that our
null hypothesis was false.
Step 3:
The shaded region is the critical
region i.e. If we observe a
sample mean in this area we will
reject the null hypothesis.
124
127.29
c. Describe the sampling distribution of x if = 128 (This is a “what-if” scenario. We are assuming that our true
population mean is 128 instead of the hypothesized mean. We are then going to determine the likelihood of
failing to reject a false null hypothesis given the true mean is 128).
Mean (  x ) = 128
Standard deviation (  x ) = 2 (will have the same standard deviation as before as we are assuming the
variance has not changed, only the location of the mean).
Shape: This will be a Bell-Shaped (normal) curve with a mean of 128 and a standard deviation of 2
On your graph from part a, sketch the sampling distribution of x if = 128
Steps 4 and 5.
The area under the Green section of the second curve would be the probability of having a value less than the
critical value of 127.29 and that the true mean is 128 (i.e. this area represents the type II error).
d. Find the probability of failing to reject H0 if  is actually equal to 128. That is, find the probability of making a
Type 2 error. Shade the area corresponding to this probability on your graph.
Step 6
We are now going to calculate the probability of getting a value less than 127.29 (the critical value) given
that the “true” mean is 128.
P( x < 127.29) = P( z <
.
.
.
Thus we have a 35.94% chance of accepting a false null hypothesis. The power of the test is
1-0.3594 = 0.6406
This would mean we have a 64.06% chance of rejecting a false null hypothesis.
Consider a test of H0:  = 124 versus Ha:  < 124. The test uses  = 16, =0.05 and a sample size of n = 64.
a. Describe the sampling distribution of x assuming H0 is true.
Mean (  x ) = 124 (sampling distribution mean is the same as the population mean)
Standard deviation (  x ) =
√
Shape: The sampling distribution will be Bell-Shaped (normal) with a mean of 124 and a standard deviation
of 2
Sketch the sampling distribution of x assuming H0 is true. (This is step 1 of the process of finding Type-II error)
120 122 124 126 128
b. Specify the rejection region when x is used as the test statistic. Locate the rejection region on your graph from
part a. (This is steps 2 and 3 of the process of finding Type-II error).
Step 2:
Since =0.05 and we have a one-sided test. We have a critical z-value of 1.645 (See the table of
critical values in the formula sheet notes). This means that if our observed value is more than 1.645
standard deviations below the mean, we will reject the null hypothesis. In step 2, we want to find
the corresponding value under our given normal curve (in this case, we want to find the value that
is 1.645 standard deviations below the mean of 124 where the standard deviation is 2. In the
formula sheet, I called this
.
.
So,
∗
.
In other words, if we observe a sample mean above 120.71, then we would consider that as being too
far from the hypothesized mean to have occurred by chance and that we can then conclude that our
null hypothesis was false.
Step 3:
The shaded region is the critical
region i.e. If we observe a
sample mean in this area we will
reject the null hypothesis.
120.71
124
c. Describe the sampling distribution of x if = 122 (This is a “what‐if” scenario. We are assuming that our true population mean is 122 instead of the hypothesized mean. We are then going to determine the likelihood of failing to reject a false null hypothesis given the true mean is 122).
Mean (  x ) = 122
Standard deviation (  x ) = 2 (will have the same standard deviation as before as we are assuming the
variance has not changed, only the location of the mean).
Shape: This will be a Bell-Shaped (normal) curve with a mean of 122 and a standard deviation of 2
On your graph from part a, sketch the sampling distribution of x if = 122
Steps 4 and 5.
The area under the Green section of the second curve would be the probability of having a value greater than the
critical value of 120.71 and that the true mean is 122 (i.e. this area represents the type II error.
d. Find the probability of failing to reject H0 if  is actually equal to 122. That is, find the probability of making a
Type 2 error. Shade the area corresponding to this probability on your graph.
Step 6
We are now going to calculate the probability of getting a value greater than 120.71 (the critical value)
given that the “true” mean is 122.
P( x > 120.71) = P( z >
.
.
.
Thus we have a 74.22% chance of accepting a false null hypothesis. The power of the test is
1-0.7422 = 0.2578
This would mean we have a 25.78% chance of rejecting a false null hypothesis.