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BA 275 Quantitative Business Methods. Agenda. Hypothesis Testing Elements of a Test Concept behind a Test Examples . Midterm Examination #1. Question 1 – A. Question 1 – B. Question 1 – C. Question 2. 982.5=(980+985)/2. 725. 8949.16=(94.6) 2. 61.28295. 575. 1247. 307=1247-940.
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BA 275 Quantitative Business Methods Agenda • Hypothesis Testing • Elements of a Test • Concept behind a Test • Examples
Question 2 982.5=(980+985)/2 725 8949.16=(94.6)2 61.28295 575 1247 307=1247-940 1000=965+35 56=750-694
Question 3 – A, B, and C 5k 6k 7k 8k 9k 10k 11k • Estimate the percentage of games that have between 7,000 to 10,000 people in attendance. 81.5% = 68% + 13.5% B. Estimate the percentage of games that have less than 6,000 people in attendance. 2.5% = 2.35% + 0.15% C. Estimate the percentage of games that have more than 9,000 people in attendance. 16% = 13.5% + 2.35% + 0.15%
Question 3 – D and E D. If we assume that the attendance follows a normal distribution with the same mean and standard deviation, estimate the percentage of games that have between 7,500 to 10,250 people in attendance. P(7500 < X < 10250) = P( -0.5 < Z < 2.25) = 0.9878 – 0.3085 = 0.6793 E. Again, with the same normality assumption, estimate the percentage of games that have more than 5,500 people in attendance. P( X < 5500 ) = P( Z < -2.5) = 0.0062 P( X > 5500 ) = 1 – 0.0062 = 0.9938
Question 4 • A bottling company uses a filling machine to fill plastic bottles with cola. The bottles are supposed to contain 300 milliliters (ml). In fact, the contents vary according to a normal distribution with mean m = 298 ml and standard deviation s = 3 ml. • A. What is the probability that an individual bottle contains less than 296 ml? • B. What is the probability that the mean contents of the bottles in a six-pack is less than 296 ml? P( X < 296 ) = P( Z < -0.67 ) = 0.2514
Question 5 • A questionnaire about study habits was given to a random sample of students taking a large introductory statistics class. The sample of 25 students reported that they spent an average of 110 minutes per week studying statistics. Assume that the standard deviation is 40 minutes. • Give a 90% confidence interval for the mean time spent studying statistics by students in this class. • If we wish to reduce the margin of error to only 8 minutes while keeping the confidence level at 90%, how large a sample do we need?
Central Limit Theorem (CLT) • The CLT applied to Means
Example 1 • The number of cars sold annually by used car salespeople is normally distributed with a standard deviation of 15. A random sample of 400 salespeople was taken and the mean number of cars sold annually was found to be 75. Find the 95% confidence interval estimate of the population mean. Interpret the interval estimate.
Statistical Inference: Estimation Example: s = 10,000 n = 100 What is the value of m? Population Research Question: What is the parameter value? Example: m =? Sample of size n Tools (i.e., formulas): Point Estimator Interval Estimator
Example 2: Concept behind a H.T. • 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?
Statistical Inference: Hypothesis Testing Example: s = 10,000 n = 100 Is “m > 22,000”? Population Research Question: Is a claim about the parameter value supported? Example: “m > 22,000”? Sample of size n Tool (i.e., formula): Z or T score
Before collecting data After collecting data Elements of a Test • Hypotheses • Null Hypothesis H0 • Alternative Hypothesis Ha • Test Statistic • Decision Rule (Rejection Region) • Evidence (actual observed test statistic) • Conclusion • Reject H0 if the evidence falls in the R.R. • Do not reject H0 if the evidence falls outside the R.R.
Example 2 (cont’d) • 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?
Type I and II Errors Chance of making Type I error = P( Type I error ) = a Chance of making Type II error = P( Type II error ) = b
Example 3 • The manager of a department store is thinking about establishing a new billing system for the store’s credit customers. After a thorough financial analysis, she determines that the new system will be cost-effective only if the mean monthly account is greater than $70. A random sample of 200 monthly accounts Is drawn, for which the sample mean account is $76 with a standard deviation of $30. Is there enough evidence at the 5% significance level to conclude that the new system will be cost-effective? • What if the sample mean is $68? $74?
Answer Key to the Examples • Example 1: • Example 2: the z score of the sample mean 100 is -3.9123 meaning the sample mean is almost 4 standard deviations below the null hypothesis m = 120. Strong evidence to reject H0 and to conclude that the true mean waiting time is less than 120 seconds. • Example 2 (cont’d): the z score of the sample mean 112 is only -1.56, less than 2 standard deviations below m = 120. No evidence to reject H0. • Example 3: at a = 5%, the rejection region is: Reject H0 if z > 1.645. • The z score of 76 is 2.828. Reject H0. • 68 is below the null m = 70. No evidence at all to reject H0. • The z score of 74 is 1.8856. Reject H0.