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

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?

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

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

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

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.

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.

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

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

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

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

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

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.

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.  

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

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.

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.

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.

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

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.

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?

Left-tail test: calculating  (cont.) Test statistic: H0 : m = 24 HA : m < 24 Choose  = .05 Rejection region in terms of z: z < -z.05 = -1.645 a = 0.05

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

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

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.

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)?

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

A Two – Tail Test (cont.) Rejection region t99 a/2 = 0.025 a/2 = 0.025 -ta/2 = -1.9842 ta/2 = 1.9842 Rejection region

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) = .2368 > .05 a/2 = 0.025 a/2 = 0.025 A Type II error is possible -1.19 1.19 -ta/2 = -1.9842 ta/2 = 1.9842

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. (17.09-1.50=15.59) What is (15.59), the probability of a Type II error when the true mean competitor’s bill  is $15.59?

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

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

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

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