1. Type II Error, Power and Sample Size Calculations
Chapter 23 Inference for Means: Part 2
Type II Error, Power and Sample Size Calculations

2. Hypothesis Testing for , Type II Error Probabilities (Right-tail example)
A new billing system for a department store will be cost- effective only if the mean monthly account is more than $170.
A sample of 400 accounts has a mean of $174 and s = $65.
Can we conclude that the new system will be cost effective?

3. Example (cont.) Hypotheses
The population of interest is the credit accounts at the store.
We want to know whether the mean account for all customers is greater than $170.
H0 : m = 170
HA : m > 170
Where m is the mean account value for all customers

4. Example (cont.)
Test statistic:
H0 : m = 170
HA : m > 170

5. Type II error is possible
Example (cont.)
P-value: The probability of observing a value of the test statistic as extreme or more extreme then t = 1.23, given that m = 170 is…
t399
Since the P-value > .05, we conclude that there is not sufficient evidence to reject H0 : =170.
Type II error is possible

6. Calculating , the Probability of a Type II Error
Calculating  for the t test is not at all straightforward and is beyond the level of this course
The distribution of the test statistic t is quite complicated when H0 is false and HA is true
However, we can obtain very good approximate values for  using z (the standard normal) in place of t.

7. Calculating , the Probability of a Type II Error (cont.)
We need to
specify an appropriate significance level ;
Determine the rejection region in terms of z
Then calculate the probability of not being in the rejection when  = 1, where 1 is a value of  that makes HA true.

8. Example (cont.) calculating 
Test statistic:
H0 : m = 170
HA : m > 170
Choose  = .05
Rejection region in terms of z: z > z.05 = 1.645 a
= 0.05

9. Example (cont.) calculating 
Express the rejection region directly, not in standardized terms
The rejection region with a = .05.
Let the alternative value be m = 180 (rather than just m>170)
H0: m = 170
HA: m = 180
a=.05
m= 170
m=180
Specify the alternative value under HA.
Do not reject H0

10. Example (cont.) calculating 
A Type II error occurs when a false H0 is not rejected. Suppose =180, that is, H0 is false.
A false H0…
H0: m = 170
…is not rejected
H1: m = 180
a=.05
m= 170
m=180

11. Example (cont.) calculating 
Power when =180 = 1-(180)=.9236
H0: m = 170
H1: m = 180
m=180
m= 170

12. Effects on b of changing a
Increasing the significance level a, decreases the value of b, and vice versa.
a2 >
b2 < a1 b1
m= 170
m=180

13. Judging the Test
A hypothesis test is effectively defined by the significance level a and by the sample size n.
If the probability of a Type II error b is judged to be too large, we can reduce it by
increasing a, and/or
increasing the sample size.

14. Judging the Test Increasing the sample size reduces b
By increasing the sample size the standard deviation of the sampling distribution of the mean decreases. Thus, the cutoff value of for the rejection region decreases.

15. Judging the Test Increasing the sample size reduces b
Note what happens when n increases:
a does not change,
but b becomes smaller
m=180
m= 170

16. Judging the Test Increasing the sample size reduces b
In the example, suppose n increases from 400 to 1000.
a remains 5%, but the probability of a Type II error decreases dramatically.

17. A Left - Tail Test Self-Addressed Stamped Envelopes.
The chief financial officer in FedEx believes that including a stamped self-addressed (SSA) envelope in the monthly invoice sent to customers will decrease the amount of time it take for customers to pay their monthly bills.
Currently, customers return their payments in 24 days on the average, with a standard deviation of 6 days.
Stamped self-addressed envelopes are included with the bills for 75 randomly selected customers. The number of days until they return their payment is recorded.

18. A Left - Tail Test: Hypotheses
The parameter tested is the population mean payment period (m) for customers who receive self-addressed stamped envelopes with their bill.
The hypotheses are: H0: m = 24 H1: m < 24
Use  = .05; n = 75.

19. A Left - Tail Test: Rejection Region
The rejection region:
t < t.05,74 = 1.666
Results from the 75 randomly selected customers:

20. A Left -Tail Test: Test Statistic
The test statistic is:
Since the rejection region is
We do not reject the null hypothesis.
Note that the P-value = P(t74 < -1.52) = .066.
Since our decision is to not reject the null hypothesis,
A Type II error is possible.

21. Left-Tail Test: Calculating , the Probability of a Type II Error
The CFO thinks that a decrease of one day in the average payment return time will cover the costs of the envelopes since customer checks can be deposited earlier.
What is (23), the probability of a Type II error when the true mean payment return time  is 23 days?

22. Left-tail test: calculating  (cont.)
Test statistic:
H0 : m = 24
HA : m < 24
Choose  = .05
Rejection region in terms of z: z < -z.05 = a
= 0.05

23. Left-tail test: calculating  (cont.)
Express the rejection region directly, not in standardized terms
The rejection region with a = .05.
Let the alternative value be m = 23 (rather than just m < 24)
m=24
H0: m = 24
Specify the alternative value under HA.
m= 23
HA: m = 23
Do not reject H0
a=.05

24. Left-tail test: calculating  (cont.)
Power when =23 is 1-(23)=1-.58=.42
H0: m = 24
HA: m = 23
m=24
a=.05
m= 23

25. A Two - Tail Test for 
The Federal Communications Commission (FCC) wants competition between phone companies. The FCC wants to investigate if AT&T rates differ from their competitor’s rates.
According to data from the (FCC) the mean monthly long-distance bills for all AT&T residential customers is $17.09.

26. A Two - Tail Test (cont.)
A random sample of 100 AT&T customers is selected and their bills are recalculated using a leading competitor’s rates.
The mean and standard deviation of the bills using the competitor’s rates are
Can we infer that there is a difference between AT&T’s bills and the competitor’s bills (on the average)?

27. A Two - Tail Test (cont.) Is the mean different from 17.09?
n = 100; use  = .05
H0: m = 17.09

28. A Two – Tail Test (cont.) Rejection region t99 a/2 = 0.025
a/2 = 0.025
-ta/2 =
ta/2 =
Rejection region

29. A Two – Tail Test: Conclusion
There is insufficient evidence to conclude that there is a difference between the bills of AT&T and the competitor.
Also, by the P-value approach:
The P-value = P(t < -1.19) + P(t > 1.19)
= 2(.1184) = > .05
a/2 = 0.025
a/2 = 0.025
A Type II error is possible
-1.19
1.19
-ta/2 =
ta/2 =

30. Two-Tail Test: Calculating , the Probability of a Type II Error
The FCC would like to detect a decrease of $1.50 in the average competitor’s bill. ( =15.59)
What is (15.59), the probability of a Type II error when the true mean competitor’s bill  is $15.59?

31. Two – Tail Test: Calculating  (cont.)
Rejection region
a/2 = 0.025
a/2 = 0.025
Do not reject H0
17.09
Reject H0

32. Two – Tail Test: Calculating  (cont.)
Power when =15.59 is 1-(15.59)=.972
H0: m = 17.09
HA: m = 15.59
m=17.09
m= 15.59

33. General formula: Type II Error Probability (A) for a Level  Test

34. Sample Size n for which a level  test also has (A) = 