1 BA 275 Quantitative Business Methods Hypothesis Testing Elements of a Test Concept behind a Test Examples Agenda

2 Midterm Examination #1

3 Question 1 – A

4 Question 1 – B

5 Question 1 – C

6 Question =(94.6) = = = =( )/2

7 Question 3 – A, B, and C A.Estimate the percentage of games that have between 7,000 to 10,000 people in attendance. B. Estimate the percentage of games that have less than 6,000 people in attendance. C. Estimate the percentage of games that have more than 9,000 people in attendance. 5k 6k 7k 8k 9k 10k 11k 81.5% = 68% % 2.5% = 2.35% % 16% = 13.5% % %

8 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. E. Again, with the same normality assumption, estimate the percentage of games that have more than 5,500 people in attendance. P(7500 < X < 10250) = P( -0.5 < Z < 2.25) = – = P( X < 5500 ) = P( Z < -2.5) = P( X > 5500 ) = 1 – =

9 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  = 298 ml and standard deviation  = 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 < ) =

10 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?

11 Central Limit Theorem (CLT) The CLT applied to Means

12 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.

13 Statistical Inference: Estimation Research Question: What is the parameter value? Sample of size n Population Tools (i.e., formulas): Point Estimator Interval Estimator Example:  = 10,000 n = 100 What is the value of  ? Example:  ?

14 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?

15 Statistical Inference: Hypothesis Testing Research Question: Is a claim about the parameter value supported? Sample of size n Population Tool (i.e., formula): Z or T score Example:  = 10,000 n = 100 Is “  > 22,000”? Example: “  22,000”?

16 Elements of a Test Hypotheses Null Hypothesis H 0 Alternative Hypothesis H a Test Statistic Decision Rule (Rejection Region) Before collecting data After collecting data Evidence (actual observed test statistic) Conclusion Reject H 0 if the evidence falls in the R.R. Do not reject H 0 if the evidence falls outside the R.R.

17 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?

18 Type I and II Errors Chance of making Type I error = P( Type I error ) =  Chance of making Type II error = P( Type II error ) = 

19 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?

20 Answer Key to the Examples Example 1: Example 2: the z score of the sample mean 100 is meaning the sample mean is almost 4 standard deviations below the null hypothesis  = 120. Strong evidence to reject H 0 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  = 120. No evidence to reject H 0. Example 3: at  = 5%, the rejection region is: Reject H 0 if z > The z score of 76 is Reject H is below the null  = 70. No evidence at all to reject H 0. The z score of 74 is Reject H 0.