1. Welcome to MM207 - Statistics! Unit 7 Seminar: Insert Day, Date here Insert time here ET Professor: Insert your name here Good Evening Everyone! To resize your pods: Place your mouse here. Left mouse click and hold. Drag to the right to enlarge the pod. To maximize chat, minimize roster by clicking here

2. Hypothesis Testing - Example Criminal Trial Null hypothesis: H0 = 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)

3. Hypothesis Testing 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

4. Hypothesis Tests: 3 types Null hypothesis always in the form: H0: 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

5. Determining the P-value The P-value is critical in determining if H0 should be rejected. P-value is one of three values (pg 379): –Left-tailed test: P-value = P(z < z x ) –Right-tailed test: P-value = P(z > z x ) –Two-tailed test: P-value = 2 * P(z > z x ) Note: z x is the test statistic. Rejection criteria –If P-value ≤ α, reject H0 –If P-value > α, do not reject H0

6. Calculating the Test Statistics Same procedure that we used in Unit 6 σ is known (normal distribution) z = (xbar - µ) / (σ/√n) σ is unknown (student t distribution) t = (xbar - µ) / (s/√n) with d.f. = n - 1

7. Working a Hypothesis Test Problem 1.Write down the information you know. 2.Determine the Null and Alternative hypothesis 3.Find the test statistic 1.σ is known 1.If normally distributed, use a z statistic 2.If distribution is unknown but n ≥ 30, use a z statistic 2.σ is unknown 1.If normally distributed, use a t statistic 2.If distribution is unknown but n ≥ 30, use a t statistic 4.Find the P-value based on the “tail type” and test statistic 5.If P-value ≤ α then reject H0. If P-value > α, then do not reject H0. 6.Summarize results

8. Section 7.2, problem #4 In an advertisement, a pizza shop claims that its mean delivery time is less than 30 minutes. A random sample of 36 delivery times has a sample mean of 28.5 minutes and a standard deviation of 3.5 minutes. Is there enough evidence to support the claim at α = 0.01? Use a p-value. Step 1: Write down what you know: µ = 30 (historical information) xbar = 28.5 (mean obtained from sampling) σ = 3.5 (standard deviation from historic research) n = 36 (sample size) α = 0.01 (level of significance, given by problem)

9. Section 7.2, problem #4 Step 1: Write down what you know: µ = 30 (historical information) xbar = 28.5 (mean obtained from sampling) σ = 3.5 (standard deviation obtained from historic research) n = 36 (sample size) α = 0.01 (level of significance, given by problem) Step 2: Set up Hypothesis test From the problem H0: µ ≥ 30 (the mean delivery time is more than 30 minutes) Ha: µ < 30 (the mean deliver time is less than 30, left-tailed test)

10. Section 7.2, problem #4 µ = 30 (historical information) xbar = 28.5 (mean obtained from sampling) σ = 3.5 (standard deviation from historic research) n = 36 (sample size) α = 0.01 (level of significance, given by problem) Step 3: Find test statistic. We know σ so we will use a z statistic z = (xbar - µ) / (σ/√n) = (28.5 – 30) / (-1.5/√36) = -2.57 Step 4: Find the P-value for a right-tailed test P-value = P(z < z x )= P(z < -2.57) = 0.0051

11. Section 7.2, problem #4 P-value = 0.0051 α = 0.01 Step 5: If P-value ≤ α then reject H0 0.0051 ≤ 0.01 is true We reject H0 Step 6: Summarize the results At a 1% significance level there is insufficient evidence to say that the average delivery time is less than 30 minutes.

12. Section 7.2, problem #4: Answer a)H0: µ ≥ 30 Ha: µ < 30 (left-tailed test) level of significance = α = 0.01 b)Normal distribution since n > 30 and σ ≈ s z = -2.57 c)Sketch not required P-value = 0.0051 d)Since P-value is less than α, we reject H0. The data is statistically significant at 1%. e)At a 1% significance level, there is sufficient evidence to say the average pizza delivery time is less than 30 minutes. Note: The above is only the answers. You should show more intermediate work to receive partial credit for incorrect answers.

13. Section 7.3, problem #5 (page 401) An industrial company claims that the mean pH level of the water in a nearby river is 6.8. You randomly select 19 water samples and measure the pH of each. The sample mean and standard deviation are 6.7 and 0.24, respectively. Is there enough evidence to reject the company’s claim at α = 0.05? Assume the population is normally distributed. Step 1: Write down what you know: µ = 6.8 (historical information) xbar = 6.7 (mean obtained from sampling) s = 0.24 (sample standard deviation from sample) n = 19 (sample size) α = 0.05 (level of significance, given by problem)

14. Section 7.3, problem #5 Step 1: Write down what you know: µ = 6.8 (historical information) xbar = 6.7 (mean obtained from sampling) s = 0.24 (sample standard deviation from sample) n = 19 (sample size) α = 0.05 (level of significance, given by problem) Step 2: Set up Hypothesis test From the problem H0: µ = 6.8 (pH value is 6.8) Ha: µ ≠ 6.8 (reject company’s claim, two-tailed test)

15. Section 7.3, problem #5 µ = 6.8 (historical information) xbar = 6.7 (mean obtained from sampling) s = 5.2 (sample standard deviation from sample) n = 19 (sample size) α = 0.05 (level of significance, given by problem) Step 3: Find test statistic. We do not know σ so we will use a t statistic. t = (xbar - µ) / (s/√n) = (6.7 – 6.8) / (0.24/√19) = -1.816 |-1.816| = 1.816 is to, the critical value Step 4: Find the P-value for a two-tailed test d.f. = n – 1 = 19 – 1 = 18 For d.f. = 18, we have 1.816 falling between 1.734 and 2.101. t = 1.734 has an area of 0.100 and t = 2.101 has an area of 0.050 P-value interval is: 0.050 < P-value < 0.100

16. Finding P-value interval T = 1.816 d.f. = 18

17. Section 7.3, problem #5 P-value interval: 0.050 < P-value < 0.100 α = 0.05 Step 5: If P-value ≤ α then reject H0 From the P-value interval, we know that the P-value is larger than 0.05 so we do not reject H0. Step 6: Summarize the results At a 5% significance level there is insufficient evidence to say that the mean pH is 6.8.

18. Section 7.3, problem #5: Answer a)H0: µ = 6.8 Ha: µ ≠ 6.8 (two-tailed test) level of significance = α = 0.01 b)Student’s t distribution since n < 30 and σ is unknown, distributed normal t = -1.816 c)Sketch not required P-value interval: 0.050 < P-value < 0.100 d)Since P-value is greater than α, we do not reject H0. The data is not statistically significant at 5%. e)At a 5% significance level there is insufficient evidence to say that the pH value is 6.8. Note: The above is only the answers. You should show more intermediate work to receive partial credit for incorrect answers.