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Handout 12. Chapter 8.4: Testing hypotheses about a population mean m (large samples). Example 2 . In trying a person for a crime, the jury needs to decide between one of two possibilities: Example 1.Suppose that a pharmaceutical company is concerned that the mean potency µ of an antibiotic does not meet the minimum government potency standards. They need to decide between two possibilities: –The mean potency µ does not exceed the mean allowable potency. – The mean potency µ exceeds the mean allowable potency. – The person is guilty. – The person is innocent. • To begin with, the person is assumed innocent. • The prosecutor presents evidence, trying to convince the jury to reject the original assumption of innocence, and conclude that the person is guilty. Main steps in testing statistical hypotheses. Types of Hypotheses: Two-Tailed Tests • 1.Formulate the null hypothesis and the research (alternative) hypothesis. The null hypothesis, H0:status quo, present state. Assumed to be true until we can prove otherwise. The alternative hypothesis, H1: it is called research, because it implies that some action is to be performed. We do not prove H0 rather look for evidence to support H1. – H1 will be accepted as true if we can disprove H0 Court trial: • Example 3: Is it true that the average hourly wage of construction workers in California is different from $14, which is the national average? – Answering this question is equivalent to testing the hypothesis : H0: µ=14 against H1: µ≠14 where µ is the average hourly wage of construction workers in the USA. This is an example of a two-tailed test of hypotheses. µ<14 µ=14 µ>14 Pharmaceuticals: H0: innocent H0: µ does not exceed allowed amount H1: guilty H1: µ exceeds allowed amount directions of H1 1 Intuition Types of Hypotheses: The one-tailed test The statistic used to estimate the population mean µ is the sample mean. Data provide evidence against H0 , if the value of x is larger than 8. How much larger? • Example 4: The average duration of Alzheimer’s disease from the onset of symptoms until death is 8 years. A pharmaceutical company claims to have developed a new drug which increases the expected lifetime. – Verifying the company’s claim would be equivalent to testing: H0: µ=8 against H1: µ>8 where µ is the average duration of the disease (i.e. average time before death). This is an example of the one-tailed test of hypotheses. µ=8 µ>8 µ=8 Data provide evidence against H0 if the value of is different from 14. How much difference? µ<14 direction of H1 1.Formulate the null hypothesis and the research (alternative) hypothesis. 2.Find the test statistic : 3.Find the rejection region (based on the critical values approach or p-values ): – A rule that tells us for which values of the test statistic, or for which p-values, the null hypothesis should be rejected. 4.Conclusion: – Either “Reject H0” or “Do not reject H0”, along with a statement about the reliability of your conclusion. How do you decide when to reject H0? – Depends on the significance level α, the maximum tolerable risk in making a mistake, if you decide to reject H0. – Usually, the significance level is α = .01 , α = .1or α = .05. µ=14 x µ>14 2.Define a sample based test statistic- a statistic calculated from the sample which will allow us to reject or not reject H0. Parts of a Statistical Test- A single statistic calculated from the sample which will allow us to reject or not reject H0. µ>8 • Assume that H0 is true. The sample mean x is our best estimate of µ, and we use it in a standardized form as the test statistic: Z= x − µ0 x − µ0 ≈ σ / n s/ n since x has an approximate normal distribution with mean µ0 and standard deviation σ / n . 2 Test Statistic for two –sided hypothesis • • • If H0 is true the value of x should be close to µ0, and z will be close to 0. If H0 is false, x will be much larger or smaller than µ0, and z will be much larger or smaller than 0, indicating that we should reject H0. For one- tailed : – H1:µ > µ0 or H1: µ < µ0 Rejection regions are calculated using only one tail of the sampling distribution. Two Types of Errors There are two types of errors which can occur in a statistical test. Actual Fact Guilty Jury’s Decision Innocent Actual Fact H0 true Your (Accept H0) Decision H0 false (Reject H0) Guilty Correct Error Correct Type II Error Innocent Error Correct H0 true (Accept H0) H0 false (Reject H0) Type I Error Correct Define: • In this type of reasoning we can make errors: – Type I error: reject the null hypothesis when this is true. (Indeed the null hypothesis is true but we observe a rare event, with a small probability) – Type II error: the null hypothesis is not true, but we do not reject it. • In Example4. Type I error –the drug in reality is not effective, but the sample average is much bigger than 8,we did reject H0 . Result :we put a not effective drug in a market. Type II error –this drug is effective , but based on our sample we failed to reject H0 .Result : We did not put the effective drug on the market. Significance Level-defining the critical values and rejection regions. Fix a maximum value for the probability of the first type error. This value is denoted by α and called significance level of the test. • We then look for the critical value to divide the sample space in acceptance and rejection region so that the probability of the first type error is kept equal to α. α zα α/2 zα/2 0.10 0.05 0.01 1.282 1.645 2.326 0.05 0.025 0.005 1.645 1.960 2.576 Critical value α = P(Type I error) = P(reject H0 when H0 is true) β =P(Type II error) = P(accept H0 when H0 is false) The value of a is the significance level, and is controlled by the experimenter. Sample space 8 Acceptance region:A Area = α Rejection region: R 3 : Scheme • Example -continue . The average duration of Alzheimer’s disease from the onset of symptoms until death is 8 years. A pharmaceutical company claims to have developed a new drug which increases the expected lifetime. Verifying the company’s claim would be equivalent to testing: To test H0: µ=µ0 use the test statistic Z= X − µ0 S n H ~ N(0,1) H 1. If H1: µ>µ0 the rejection region is R={ values z such that z > zα } 2. If H1: µ< µ0 the rejection region becomes R={ values z such that z <- zα }. 3. If H1: µ≠ µ0 the rejection region becomes R={ values z such that either z <- zα/2 or z > zα/2 }. Find the rejection region: α = 0.05, find zα such that p(Z> zα )=0.05. zα = 1.645 Decision : Reject H 1.645 Acceptance o if z calc 0 > z α = 1.645 z=8 Rejection Since the value of the test statistic lies in the rejection region, the conclusion is to reject H0 and conclude that this drug is effective and increases the expected lifetime. 1 We test : µ = 8 : µ > 8 have a sample X − µ 0 s / n calculate 10 − 8 z calc = 2 / 64 Z ( or µ ≤ 8 ) n = 64, x = 10; S 2 = 4 = = 8 Acceptance/Rejection-another method P- value is the probability ,calculated under Ho that the test statistic takes a value equal to or more extreme than the value observed. -A large p-value means that the event we observe is highly likely, if Ho is true, so there is nothing against the null hypothesis. Do not reject H0 - A small p-value means that the event we observe is very unlikely if the null hypothesis is true. Reject H0. Conclusions: we reject the null hypothesis If the p-value is smaller than the preassigned value a, If a is not given – use a=0.05. 4 Example 3 –continue Is it true that the average hourly wage of construction workers in California is different from $14, which is the national average? We then use the following reasoning: If H0 is true, how likely is it to observe a value of the sample mean equal to or larger than 10? We want to compute the probability of this event: From a sample of 49 individuals, we compute that the sample mean is 14.5 and the standard deviation is 2. Area = α/2 1) Area = α/2 p( X 14 | H0) 10 - 8 ) 2 64 ≅ 0 p(Z ≥ p(Z ≥ Rejection Acceptance Rejection region region region Critical values ≥ 10 8) = So, data provide evidence against the null hypothesis. The conclusion is: reject H0 and accept H1and conclude that the drug increases the expected lifetime. H0 : µ = 14 H1 : µ ≠ 14 test 3) 4) Let α = 0.05, find zα/2 such that P(Z> zα/2 )=0.025. Zα/2 = 1.96 Reject Ho if z calc >1.96 or Z calc <-1.96 x − µ0 Z= s/ n 14.5 − 14 z calc= = 1.75 2 / 49 -1.96 Rejection 1.96 Acceptance 5) Do not reject Ho and conclude that the average wage is $14. p - value = P (Z > 1.75) = (1 − 0 . 9599 ) + 0 . 0401 + P(Z < - 1.75) = = 0 . 0802 or p − value p − value = 2 P ( Z > 1 . 75 ) = 0 . 0802 > α = 0 . 05 we do not reject H o at α = 0 . 05 Rejection 5 p-value approach Review: Steps in Hypothesis Testing 1. Specify the population value of interest. 2. Formulate the appropriate null and alternative hypotheses. 3. Specify the desired level of significance. – Also called observed level of significance 4. Determine the rejection region. 5. Obtain sample evidence and compute the statistic. • p-value: Probability of obtaining a test statistic more extreme ( ≤ or ≥ ) than the observed sample value given H0 is true test 6. Reach a decision and interpret the result. – Smallest value of α for which H0 can be rejected • Compare the p-value with α – If p-value < α , reject H0 – If p-value ≥ α , do not reject H0 Hypothesis on a population mean. Suppose the hypotheses are about the parameter m-population mean . H0: m=m0 The test statistic is x − µ Z = 0 σ n and the rejection region is given by : 1) If H1; m>m0 Reject H0 when zcalc>zα p-value =P(Z > zcal) 2) If H1; m<m0 Reject H0 when zcalc < -zα p-value =P(Z < zcal) 3) If H1; m∫m0 Reject H0 when | zcalc| > zα/2 it means: p-value =P(|Z| >zcal) zcalc <-zα/2 or zcalc > zα/2 , Reject H0 when p-value is small , let’s say p-value <a ,or p-value <0.05. Example 5 A phone industry manager thinks that customer monthly cell phone bill have increased, and now average over $52 per month. The company wishes to test this claim. (Assume σ = 10 is known) Form hypothesis test: H0: µ ≤ 52 or =52 the average is not over $52 per month H1: µ > 52 the average is greater than $52 per month (i.e., sufficient evidence exists to support the manager’s claim) 6 Test statistic, zcalc Example: Find Rejection Region Obtain sample evidence and compute the test statistic Suppose a sample is taken with the following results: n = 64, sample mean = 53.1 (σ=10 was assumed known) – Then the test statistic is: Suppose that α = .10 is chosen for this test Find the rejection region: Reject H0 Z= x − µ 53.1 − 52 = = 0.88 σ 10 64 n α = .10 Do not reject H0 0 zα=1.28 Reject H0 Reject H0 if z > 1.28 Example: Decision Reach a decision and interpret the Reject H0 result: p -Value Solution Calculate the p-value and compare to α p-value = .1894 P( X ≥ 53.1 | µ = 52.0) α = .10 Do not reject H0 0 1.28 z = .88 Reject H0 Do not reject H0 since z = 0.88 ≤ 1.28 i.e.: there is not sufficient evidence that the mean bill is over $52 (continued) Reject H0 α = .10 0 Do not reject H0 1.28 z = .88 Reject H0 ⎛ ⎞ ⎜ 53.1 − 52.0 ⎟ = P⎜ Z < ⎟ 10 ⎜ ⎟ 64 ⎠ ⎝ p − value = P(Z ≥ 0.88) = = .1894 Do not reject H0 since p-value = .1894 > α = .10 7 Example 6 • How much time do computer users spend on the web? An Internet provider claims that the time spent per week has a normal distribution with mean µ=13 hours per week and standard deviation σ=5.2 hours. In a sample of 25 customers, the average time was found to be 10 hours. Do the data support the claim that the average time spent by the population of computer users is less than 13 hours per week? Use α=0.1. Example 7 • The manager of a health maintenance organization has set as target that the mean waiting time of patients will not exceed 30 minutes, and he wishes to check whether this target is met. In a sample of 22 patients, the average waiting time was 38 minutes. It can be assumed that the waiting time in the population has normal distribution with standard deviation σ=10, and that α=0.01. • State the null and alternative hypothesis below. • State the test statistic to use and indicate the rejection region . • Does the data provide evidence against the target? • Describe what a type I error would be and what the consequences would be. 8