6/4/10 Lecture 17 1STOR 155 Introductory StatisticsLecture 17: Hypothesis TestingSection 6.2The UNIVERSITY of NORTH CAROLINAat CHAPEL HILL6/4/10 Lecture 17 2• Point estimation– to estimate a parameter (quick and easy)• Confidence interval– to estimate a parameter with measures for reliability and accuracy attached• Hypothesis testing– hypothesis: a statement about the parameters– to assess whether the data provide enough evidence for some claim about the populationProcedures for statistical inference6/4/10 Lecture 17 3Confidence Interval• Point estimate with margin of error• Confidence interval for a population mean– Assumption: the population variance is known– Confidence level C determines z*nzxnzx**,6/4/10 Lecture 17 4Hypothesis Testing• Sometimes you may not be interested inestimating the parameter value …• Rather, you have some claim (belief) about the parameter and you want to see whether the data support the claim or not, i.e. to choose between two decisions:``support’’ versus ``contradict’’6/4/10 Lecture 17 5“Biased” one-Euro Coin?• A group of statistics students spun the Belgian one-Euro coin 250 times, and heads came up 140 times.• p = P(H) in each spin• Claim: the coin is biased (more specifically, p is greater than 0.5)6/4/10 Lecture 17 6• Two hypotheses– H0: the null hypothesis • the statement of “no effect” or “no difference”• the statement we try to find evidence against– Ha: the alternative hypothesis• the statement we hope or suspect to be true• Usually Hais based on a ``hint’’ from samplesConcepts of Hypothesis Testing6/4/10 Lecture 17 7“Biased” one-Euro Coin? (continued)• Sample: 140 heads among 250 spins of a Belgian one-Euro coin (a hint)• p = P(H) in each spin• H0: p = 0.5 vs Ha: p > 0.5.– one-sided• H0: p = 0.5 vs Ha: p ≠ 0.5.– two-sided6/4/10 Lecture 17 8• A sprinkler system maker claims that the true average system-activation temperature is 130o. A sample of 9 systems is tested, and yields an average activation temperature of 131.08o. • If the distribution of activation temperatures is normal with = 1.5o, does the data contradict the claim?• The population consists of all sprinkler systems made.• We want to show that the mean activation temp is different from 130. Ha: m 130.• The null hypothesis must specify a single value of the parameter m. H0: m= 130.• How do we proceed ?Sprinkler6/4/10 Lecture 17 9Specify a test statistic• A test is based on a statistic, often a point estimate of the parameter that appears in the hypotheses• Values of the estimate far from the parameter value in H0give evidence against H0.• Hadetermines which direction will be counted as “far from the parameter value”.• Using standardization, the test statistic usually has the formz = (estimate - hypothesized value) / (standard deviation of the estimate)6/4/10 Lecture 17 10• A sprinkler system maker claims that the true average system-activation temperature is 130o. A sample of 9 systems is tested, and yields an average activation temperature of 131.08o. • If the distribution of activation temperatures is normal with = 1.5o, does the data contradict the claim?• The test statistic isSprinkler (continued)16.25.008.19/5.113008.131z6/4/10 Lecture 17 11• Definition: the probability of observing a test statistic value as extreme or more extreme than the actually observed value, given that H0is true. – “extreme” means “far from what we would expect from H0”.• P-value indicates how strong the statistical evidence is against the null hypothesis.– the smaller a P-value, the stronger evidence against H0P-value6/4/10 Lecture 17 12• If the P-value is less than 1%, there is overwhelming evidence against the null hypothesis.• If the P-value is between 1% and 5%, there is strong evidence against the null hypothesis.• If the P-value is between 5% and 10% there is weak evidence against the null hypothesis.• If the P-value exceeds 10%, there is no evidenceagainst the null hypothesis.Implication of P-value (rule of thumb)6/4/10 Lecture 17 13Significance Level and Statistical Significance• We need to make a conclusion after carrying out a hypothesis test. What do we conclude?• We compare the P-value with a fixed value that we regard as decisive. • This amounts to deciding in advance how much evidence against H0we require in order to reject H0. • The decisive value is called the significance level of the test. It is denoted by and the corresponding test is called a level test.Statistical Significance: If the P-value , we say that the data are statistically significant at level.6/4/10 Lecture 17 14 and P-value• P-value and significance level :– Reject H0if the P-value – Do not reject H0if the P-value >.• When is the evidence against H0stronger? – large P-value or small P-value? – The smaller the P-value, the stronger the evidence against H0.• When is it easier to reject H0? – large or small ?– We need a lot more evidence to reject H0for small than for large .6/4/10 Lecture 17 154 steps in hypothesis testing• Define the hypotheses, and identify the significance level given in the problem.• Find the value of the test statistic.• Calculate the P-value based on the data.• State your conclusion – Reject the null hypothesis if the P-value ; if the P-value >, the data do not provide sufficient evidence to reject the null hypothesis.6/4/10 Lecture 17 166/4/10 Lecture 17 17Sprinkler (continued)• Test the hypotheses: H0: m= 130 vs. Ha: m 130 at significance level = 0.01 ?• Test statistic• P-value: 2 P(Z 2.16) = 0.0308 > 0.01• H0is not rejected at significance level 0.01• Question: What is the 99% confidence interval for the activation temperature m?.16.25./08.1)9/5.1/()13008.131()//()(0 nxzm6/4/10 Lecture 17 18CI & 2-Sided Tests• A level 2-sided test accepts (or rejects) H0: m= m0exactly when the value m0falls inside (or outside) a level 1 -confidence interval for m.• Confidence interval can be used to test hypotheses.– Calculate the 1 -level confidence interval …• If m0falls within the interval, do not reject the null hypothesis, • Otherwise, reject the null hypothesis.6/4/10 Lecture 17 19• In a discussion of SAT scores, someone comments: “Because only a
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