1. 1 BA 275 Quantitative Business Methods Quiz #3 Statistical Inference: Hypothesis Testing Types of a Test P-value Agenda

2. 2 Central Limit Theorem (CLT) The CLT applied to Means

3. 3 Example 1 How much time do executives spend each day reading and sending e-mail? A survey of 162 executives was conducted and the mean time (in minutes) was 63.6975 minutes. Assume that the population std is 18.9403. Can we infer that the mean amount of time spent by all executives reading and sending e-mail exceeds 60 minutes? Assume 5% significance level.

4. 4 Example 2 How much time do executives spend each day reading and sending e-mail? A survey of 162 executives was conducted and the mean time (in minutes) was 63.6975 minutes. Assume that the population std is 18.9403. Can we infer that the mean amount of time spent by all executives reading and sending e-mail is different from 60 minutes? Assume 5% significance level.

5. 5 Example 3 How much time do executives spend each day reading and sending e-mail? A survey of 162 executives was conducted and the mean time (in minutes) was 63.6975 minutes with a standard deviation of 18.9403. At 5% significance level, we concluded that the mean amount of time exceeds 60 minutes. By how much?

6. 6 The p-Value Approach (textbook, p.386) The p-value is the probability, under the assumption that H 0 is true, of obtaining a test statistic as or more extreme than the one actually obtained from the data. (alternative definition) The p-value is the smallest value of  that would lead to the rejection of H 0. The smaller the p-value, the stronger the evidence against H0 provided by the data. Compare the p-value to the significance level .

7. 7 Example 1 (cont’d) How much time do executives spend each day reading and sending e-mail? A survey of 162 executives was conducted and the mean time (in minutes) was 63.6975 minutes. Assume that the population std is 18.9403. Can we infer that the mean amount of time spent by all executives reading and sending e-mail exceeds 60 minutes? Assume 5% significance level. Calculate the p-value.

8. 8 Example 2 (cont’d) How much time do executives spend each day reading and sending e-mail? A survey of 162 executives was conducted and the mean time (in minutes) was 63.6975 minutes. Assume that the population std is 18.9403. Can we infer that the mean amount of time spent by all executives reading and sending e-mail is different from 60 minutes? Assume 5% significance level. Calculate the p-value.

9. 9 Example 4 A bank has set up a customer service goal that the mean waiting time for its customers will be less than 2 minutes. The bank randomly samples 30 customers and finds that the sample mean is 100 seconds. Assuming that the sample is from a normal distribution and the standard deviation is 28 seconds, can the bank safely conclude that the population mean waiting time is less than 2 minutes? Find the p-value.

10. 10 Example 5 A bank has set up a customer service goal that the mean waiting time for its customers will be less than 2 minutes. The bank randomly samples 30 customers and finds that the sample mean is 112 seconds. Assuming that the sample is from a normal distribution and the standard deviation is 28 seconds, can the bank safely conclude that the population mean waiting time is less than 2 minutes? Find the p-value.

11. 11 Answer Key to the Examples Used Example 1. H 0 :  = 60 vs. H a :  > 60. Rejection region: reject H 0 if z > 1.645. Given z = 2.48, the statistical conclusion is to reject H 0. Example 2. H 0 :  = 60 vs. H a :  ≠ 60. Rejection region: reject H 0 if z > 1.96 or z < -1.96. Given z = 2.48, the conclusion is to reject H 0. Example 3. 63.6975 ± 1.96 (18.9403/sqrt(162)) Example 1 (cont’d): p-value = 1 – 0.9934 = 0.0066. Example 2 (cont’d): p-value = 2 × (1 – 0.9934) = 0.0132. Example 3: p-value ≈ 0.0000. Example 4: p-value = P( z < -1.56 ) = 0.0594