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BA 275 Quantitative Business Methods

BA 275 Quantitative Business Methods. Agenda. Quiz #3 Statistical Inference: Hypothesis Testing Types of a Test P-value. Central Limit Theorem (CLT). The CLT applied to Means. Example 1.

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BA 275 Quantitative Business Methods

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

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

  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. 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. 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. The p-Value Approach • (textbook, p.386) The p-value is the probability, under the assumption that H0 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 a that would lead to the rejection of H0. • The smaller the p-value, the stronger the evidence against H0 provided by the data. • Compare the p-value to the significance level a.

  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. 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. 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. 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. Answer Key to the Examples Used • Example 1. H0: m = 60 vs. Ha: m > 60. Rejection region: reject H0 if z > 1.645. Given z = 2.48, the statistical conclusion is to reject H0. • Example 2. H0: m = 60 vs. Ha: m≠ 60. Rejection region: reject H0 if z > 1.96 or z < -1.96. Given z = 2.48, the conclusion is to reject H0. • 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

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