Welcome to MM207 Unit 7 Seminar Dr. Bob Hypothesis Testing and Excel 1

Hypothesis Testing - Example Criminal Trial Null hypothesis: Ho = defendant is not-guilty Alternative hypothesis: Ha = person is guilty Procedure We assume Null hypothesis is true. The defendant is not-guilty until we prove otherwise. Evidence is presented and we then decide whether to: –Reject the Null hypothesis (person is guilty) –Do not reject the Null hypothesis (not enough evidence to reject, but it doesn’t mean it is true) 2

Hypothesis Testing Decision Errors Type I error Reject a true Null hypothesis (i.e. innocent person found guilty) α = alpha = probability of Type I error Type II error Do not reject a false Null hypothesis (i.e. guilty man goes free) β = beta = probability of Type II error 3

Hypothesis Tests: 3 types Null hypothesis always in the form: Ho: µ = k (mean equals a certain value) Alternative hypothesis can take 3 forms: Ha: µ k (right-tail, actual mean is greater than stated value) Ha: µ ≠ k (two-tail, actual mean is not equal to stated value) The language of the problem will tell you which alternative hypothesis you need to use. This takes practice 4

Interpreting the P-value The P-value is critical in determining if Ho should be rejected. –If P-value less than α, reject Ho –If P-value greater α, do not reject Ho 5

Section 7.2, Question 33 [page 391] In Illinois, a random sample of 85 eighth grade students has a mean score of 282 with a standard deviation of 35 on a national mathematics assessment test. The test result prompts a state school administrator to declare that the mean score for the state’s eighth graders on the examination is more than 275. At α = 0.04,is there enough evidence to support the administrator’s claim? 6

Question 33 continued State the Null and Alternative Hypotheses Ho: µ ≤ 275; please note that the equality (=) is ALWAYS in the Null) Ha: µ > 275 (the state average greater than 275). This is the Claim! Determine the Alpha level (given in this problem = 0.04) Calculate the Test Statistic (I am going to use Excel!) Interpret the Results 7

Unit 7 Excel Template 8

Results from Z-Test Mean 9 Compare these results to the Textbook Since the one-tail p-value is less than alpha (0.04) we reject the null and conclude that there is enough evidence to support the administrator’s claim that it is greater than 275.

Section 7.3, Example 4 (page 400) A used car dealer says that the mean price of a 2005 Honda Pilot LX is at least $23,900. You suspect that this claim is incorrect and find that a random sample of 14 similar vehicles has a mean price of $23,000 and a standard deviation of $1113. Is there enough evidence to reject the dealer’s claim at α = 0.05? Assume the population is normally distributed. 10

Section 7.3, Example 4 continued State the Null and Alternative Hypotheses Ho: µ ≥ $23,900; This is the Claim! [notice it can be either the null or the alternative depending on the problem. Ha: µ < $23,900 Determine the Alpha level (given in this problem = 0.05) Calculate the Test Statistic (I am going to use Excel again!) Interpret the Results 11

Results from the t-test Mean 12 Compare these results to those on page 400 We can either use the one-tail p-value (.0049 <.05) or the rejection region method (-3.03 < ) to conclude that we reject the claim and conclude that the mean price is less than $23,900.

Section 7.4, Example 1 (page 408) A research center claims that less than 20% of Internet users in the United States have a wireless network in their home. In a random sample of 100 adults, 15% say they have a wireless network in their home. At α = 0.01, is there enough evidence to support the researcher’s claim? 13

Section 7.4, Example 1 continued State the Null and Alternative Hypotheses Ho: p ≥ 0.20; Ha: p < 0.20; This is the Claim Determine the Alpha level (given in this problem = 0.01) Determine if np ≥ 5 and nq ≥ 5 so that you can use the z-test Calculate the Test Statistic (I am going to use Excel again!) Interpret the Results 14

Results from the z-test Proportion 15 Compare these results to those on page 408 We can either use the one-tail p-value (.1056 >.01) or the rejection region method (-1.25 > ) to conclude that we cannot reject the claim that the proportion is 20%.