Welcome to MM207 - Statistics!Unit 7 Seminar

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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)

Hypothesis Testing Errors

• Type I errorReject a true Null hypothesis (i.e. innocent person found guilty)α = alpha = probability of Type I error

• Type II errorDo not reject a false Null hypothesis (i.e. guilty man goes free)β = beta = probability of Type II error

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 (left-tail, actual mean is less than stated value)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

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 < zx)– Right-tailed test: P-value = P(z > zx)– Two-tailed test: P-value = 2 * P(z > zx)

Note: zx is the test statistic.

• Rejection criteria– If P-value ≤ α, reject H0– If P-value > α, do not reject H0

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

Working a Hypothesis Test Problem

1. Write down the information you know.

2. Determine the Null and Alternative hypothesis

3. Find the test statistic1. 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

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)

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 testFrom 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)

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 statisticz = (xbar - µ) / (σ/√n) = (28.5 – 30) / (-1.5/√36) = -2.57

• Step 4: Find the P-value for a right-tailed testP-value = P(z < zx)= P(z < -2.57) = 0.0051

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 resultsAt a 1% significance level there is insufficient evidence to say that the average delivery time is less than 30 minutes.

Section 7.2, problem #4: Answer

a) H0: µ ≥ 30Ha: µ < 30 (left-tailed test)level of significance = α = 0.01

b) Normal distribution since n > 30 and σ ≈ sz = -2.57

c) Sketch not requiredP-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.

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)

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 testFrom the problem

H0: µ = 6.8 (pH value is 6.8)Ha: µ ≠ 6.8 (reject company’s claim, two-tailed test)

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 testd.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

Finding P-value interval

• T = 1.816• d.f. = 18

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 resultsAt a 5% significance level there is insufficient evidence to say that the mean pH is 6.8.

Section 7.3, problem #5: Answer

a) H0: µ = 6.8Ha: 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 normalt = -1.816

c) Sketch not requiredP-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.