Significance Testing in Statistics: Hypothesis Testing with z-Test and t-Test - Prof. Rick, Study notes of Statistics

Download Significance Testing in Statistics: Hypothesis Testing with z-Test and t-Test - Prof. Rick and more Study notes Statistics in PDF only on Docsity! Chapter 8 Statistical Inference: Significance Tests About Hypotheses 8.1 What are the Steps for Performing a Signif- icance Test? A. • Definition: Hypothesis, page 368 In statistics, a hypothesis is a statement about a population, usually of the form that a parame- ter (such as µ or p) takes a particular numerical value or falls in a certain range of values. Definition: Signficance Testing is major form of inferential statistics. It is a method of using data to summarize the evidence about a hypothesis. • Five Steps of Significance Testing (see p 371) – Assumptions Specify variable and parameter. Assumptions commonly pertain to method of data produc- tion (randomization), sample size, shape of population distribution 1 2 – Hypotheses Null hypothesis, H0, statement that the pa- rameter takes a particular value Alternative hypothesis, Ha, states that the pa- rameter falls in some alternative range of val- ues. – Test Statistic Measures distance between point estimate of parameter and its null hypothesis value, usu- ally by the number of standard errors between them – P value Presume Ho to be true. The P-value is the probability that the test statistic takes the ob- served value or a value more extreme (in the direction of Ha. Smaller P-values represent stronger evidence against Ho. – Conclusion Report and interpret the P-value in the con- text of the study. Based on the P-value, make a decision about Ho is one is needed. If a deci- sion is needed, reject presumption of Ho being true and conclude Ha true if P − value ≤ α; do not reject presumption of H0 or do not con- clude Ha true is P − value > α. α is a small value, such as 0.05 or 0.01, prescribed by the researcher or some other interested party. 5 initiatives on the topic of legalized gambling have ap- peared on state ballots in recent years. Suppose that a political candidate has decided to support legaliza- tion of casino gambling if he is convinced that more than than two thirds (2/3 = 0.667) of U.S. adults approve of casino gambling. USA Today (June 17, 1999) reported the results of a Gallup poll in which 1523 adults (selected at random from households with telephones) were asked whether they approved of casino gambling. The number in the sample who approved was 1035. Does the sample provide con- vincing evidence that more than two-thirds approve? Use a significance level α = 0.05. • Assumptions – The data is binary (success = adult approves or failure = adult does not approve) – 1523 adults were selected at random – n = 1523 is sufficiently large (at the null hy- pothesis value of p) to ensure that sample dis- tribution of variable p̂ = is approximately nor- mal. Rule of Thumb: expected number of suc- cess and failures, at null hyp. value of p, are both at least 15. True here since 1523(0.667) = 1015.8 1523(1-.667) = 507.2, are both at least 15. • Hypotheses – Let p = population proportion of adult who approve of legalization of casino gambling – Ho : p = 2/3 = 0.667 (Proportion approving in population is 2/3) 6 – Ha : p > 2/3 = 0.667 (More than 2/3 of pop- ulation approves) • Test Statistic – Let p̂ be the random variable, sample propor- tion of a random sample 1523 adults who ap- prove – Test Statistic z = p̂−0.667√ 0.667(1−0.667) 1523 • P value – Suppose that H0 is true, i.e. the population approving is 2/3 = 0.667. The z test statistic has an approximate stan- dard normal distribution when Ho is true – Calculated p̂ = 1035/1523 = 0.680 Calcuated z = 0.680−0.667√ 0.667(1−0.667) 1523 Calculated z = 0.0130.012 = 1.08 – P-value = P [z ≥ 1.08] = 1 − 0.8599 = 0.1401 • Conclusion, including interpretation of P-value Interpretation: There is about a 14% chance (P = 0.1401) of obtaining a sample proportion like 0.680 or higher of adults approving (or a differ- ence like 0.013) from a population with propor- tion equal to 0.667 due solely due random/sampling variation. The P-value is not smaller than our significance level of 0.05, so there is not convinc- ing evidence against the population proportion approving being equal to 2/3, or there is not con- vincing evidence that the population approving is greater than 2/3. 7 3. Example 3 (One sided significance test) In December 2003 a county-wide water conservation campaign was conducted in a particular county. In January 2004 a random sample of 500 homes was selected, and water usage was recorded for each home in the sample. The county supervisors want to know whether the data support the claim fewer than half the households in the county reduced water consumption. Suppose that the sample proportion that reduced water con- sumption was 0.440. Use a significance level of 0.01. • Assumptions – The data is binary (success = reduced water consumption or failure = did not reduce water consumption) – 500 homes were selected at random – n = 500 is sufficiently large (at the null hy- pothesis value of p) to ensure that sample dis- tribution of variable p̂ is approximately nor- mal. Rule of Thumb: expected number of success and failures, at null hyp. value of p, are both at least 15. True here since 500(0.5) = 250 500(1-.5) = 250, are both at least 15. • Hypotheses – Let p = population proportion of homes that reduced water consumption – Ho : p = 0.5 (Exactly 0.5 of homes in pop. reduced water consumption) – Ha : p < 0.5 (Fewer than 0.5 of homes in pop. reduced water consumption) 10 – Ha : p 6= 0.5 (Proportion of overweight indi- viduals differs from national proportion) • Test Statistic – Let p̂ be the variable, sample proportion of overweight individuals in a random sample of 600 from her state – Test Statistic z = p̂−0.71√ 0.71(1−0.71) 600 • P value – Suppose that H0 is true, i.e. the population of individuals in the state that are overweight is 0.71, the same as the national rate. The z test statistic then has an approximate standard normal distribution when Ho is true – Observed p̂ = 450/600 = 0.75 Calcuated z = 0.75−0.71√ 0.71(1−0.71) 600 Calculated z = 0.040.019 = 2.11 – P-value = Two-tailed probability P-value = P [z ≥ 2.11orz ≤ −2.11] P-value = 2P [z ≤ −2.11] = 2(0.0174) = 0.0348. • Conclusion, including interpretation of P-value Interpretation: There is about a 3.5% chance (P = 0.0348) of obtaining a sample proportion as far away as 0.04 in either direction from a population proportion of 0.71 due solely to ran- dom/sampling variation. While this probability is rather small it is not small enough according to our criterion of 0.01. That is P-value > 0.01 and 11 thus there is not quite enough evidence based on this sample to conclude that the proportion of all individuals in the her state differs from the national proportion of 0.71. 5. Summary: Steps of a Significance Test for a Popula- tion Proportion p See page 381 6. Some things to know: • The P-value is a measure of how likely it is to obtain the kind of data that we got due solely to random/sampling/chance variation from a true null hypothesis. So small P-values would provide evidence against the null hypothesis and in favor of the alternative hypothesis • The significance level or α level of a test, a small number, is a prescription for how small the P- value has to be before we say we have convinc- ing evidence against the null hypothesis. Typical values of α are 0.05 or 0.01. • Thus if P − value ≤ α then there is convincing evidence against the null hypothesis and in favor of the alternative hypothesis. We say that we reject the null hypothesis and accept the alter- native hyp. If P − value > α then there is not enough evidence against the presumed truth of the null hypothesis. Then we say that we do not reject (the presumed) truth of the null hypothe- sis. 12 • A decision “Do not reject the null hypothesis” DOES NOT MEAN ACCEPT THE NULL HY- POTHESIS. It means that the null hypothesis is plausible but the alternative might be plausible as well • A decision to “Accept the alternative hypothesis” DOES MEAN THAT the alternative hypothesis is plausible and the null hypothesis is not. 7. How do we Decide between a One-sided and a Two- sided Test? (page 384, textbook) 8.3 Significance Tests about Means 1. General Form of Significance Test, page 392 2. Example: A study conducted by researchers at Penn- sylvania State University investigated whether time perception, a simple indication of a person’s abil- ity to concentrate, is impaired during nicotine with- drawal. The study results were presented in the paper “Smoking Abstinence Impairs Time Estima- tion Accuracy in Cigarette Smokers”. After a 24- hr smoking abstinence, 18 smokers were asked to estimate how much time had passed during a 45- second period. Suppose the resulting data on per- ceived elapsed time (in seconds) are as shown: 69 65 72 73 59 55 39 52 57 56 50 70 47 56 45 70 64 53 The researchers wanted to determine whether smok- ing abstinence had a negative impact on time per- 15 a poor-on-time record decides to offer its employees a bonus if, in an upcoming month, the airline’s pro- portion of ontime flights exceeds the industry rate of 0.77. Let p be the true proportion of the airline’s flights that are on time during the month of interest. A random sample of flights might be selected and used as a basis for testing Ho : p = 0.77 versus Ha : p > 0.77. 3. Example(Slowing the Growth of Tumors) Researchers at the National Cancer Institute announced plans to begin studies of a cancer treatment thought to slow the growth of tumors. Having tested the treatment on only 13 patients, the researchers had very little information, but they stated that the treat- ment appears to be less toxis than standard chemother- apy treatments. An experiment to study the treat- ment more extensively was reported as being in the planning stages. Let µ denote the true (population) mean growth rate of tumors for patients receiving the new treatment. Data resulting from the planned experiments can be used to test Ho : µ = mean growth rate of tumors without treat- ment Ho : µ < mean growth rate of tumors without treat- ment 4. The significance level is the Probability of a Type I Error. 16 As the probability of a Type I Error goes down, the P(Type II error) goes up. 8.5 Limitations of Signficance Tests 1. Statistical Signficance Does Not Mean Practical Sig- nificance 2. Significance Tests are Less Useful than Confidence Intervals 3. Misinterpretaions of Results of Significance Tests • “Do Not Reject Ho” does not mean “Accept Ho” • Statistical significance does not mean practical significance • The P-value cannot be interpreted as the proba- bility that that Ho is true. • It is misleading to report results only if they are “statistically significant” • Some tests may be statistically significant just by chance.