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1 Hypothesis (Significance) Tests About a Proportion Example 1 The standard treatment for a disease works in of all patients. A new treatment is proposed. Is it better? (The scientists who created it claim it is. You as an advocate for a patient or sales personnel for the company that eventually would market it must be more skeptical. Where s the data?) An initial clinical trial of n = 100 patients (of similar general health) is conducted: 77 people are cured. Assume the selection of patients is random. Test appropriate hypotheses at the 1% significance level. p = proportion of all patients who are cured (probability a single patient is cured; a parameter) No one will know with high precision about the value of p until many thousands of people have used the treatment. X = the number of patients in the initial trial who are cured. pˆ = X/n = the sample proportion ( pˆ is the statistic estimating p) To use the methods laid out in Textbook Steps [TS] 1 8 on page 406 the situation must meet the requirements (see the top of page 413) 1. The sample observations are a random sample The population is at least 20 times the sample size (or sampling with replacement) The values of np and n(1 p) must both be at least 10 when p from the null hypothesis is used. 2 TS Formal statement of statistical hypotheses: H 0 : H 1 : This is a -tailed test. Outcomes favorable to the alternative hypothesis are to the of We will either reject H 0 thereby endorsing H 1 as truth or fail to reject H 0 (leaving the question essentially up in the air). TS 4 The significance level (for the test) is = If our data is of the kind that occurs no more than 1% of the time when H 0 is true, we ll reject H 0. You can anticipate an answer by constructing a confidence interval. However: for an alternative hypothesis that is one-tailed (like this one) you must adjust the confidence level so that the appropriate probability (the significance level) is the area in one tail. Significance level (in one the right tail only) = Appropriate confidence: % Confidence Interval: pˆ = E = < p < Indicated decision: TS 5 Because the claim is about a proportion, and the requirements are met, we use the sample proportion, which has (approximately) a Normal distribution, to test the claim. TS 6 The test statistic uses p from H 0, and is pˆ p Z p 1 p n 1 The textbook states The conditions for a Binomial distribution are satisfied. This is equivalent to what is presented on this sheet. (Essentially what is stated here defines what constitutes a Binomial distribution. The both np and n(1 p) at least 5 requirement guarantees that the Binomial distribution under consideration can be approximated with good accuracy with a Normal distribution which is what this procedure makes use of.) 2 Your textbook uses at least 5 here. 10 is better.

2 TS 6 TRAD 3 Test stat; critical value(s); critical region. TS 6 PV 4 Test stat; P-value. TS 7 TRAD Decide: Reject H 0 or not? (Is the test stat in critical region?) TS 7 PV Decide: Reject H 0 or not? (Is P-value?) (Your two decisions cannot differ.) TS 8 Conclusion (simple; nontechnical: you may NOT use any of these terms: null, alternative, hypothesis, reject, accept (never use this word, even in technical talk, when doing hypothesis testing), alpha, significance level. See page 403 the two bottom branches. (We ll take it for granted the answer to the 1 st question is No. ) Professional talk. [B students.] This is underemphasized in the text; it appears on page 7 and then is ignored. When H 0 is rejected, a difference (or differences in more complicated studies) are said to be statistically significant. Since the determination depends on the level (people in the know recognize this to mean significance level in testing applications) a statement about statistical significance is generally accompanied by the level of the test. [A students.] Interpreting the P-value. The P-value is a probability. (See the top of page 400.) For statistical inference (intervals / tests) the units are samples and the variable is a statistic. 3 Using the Traditional method (right flowchart on page 406). 4 Using the P-value method (left flowchart on page 406).

3 Example 2 Mississippi Department of Motor Vehicle (DMV) records indicate that of all vehicles undergoing emissions testing during 2009, 70% passed on the first try. Late in 2009 the state implemented a program designed to encourage preventative service to improve emissions compliance. Does this program work? Test at the 5% level. Of 126 randomly selected reports on emissions, 95 cars pass on the first try. TS 1-2-3: Hypotheses H 0 : H 1 : TS 4: Significance level = TS 5: Because the claim is about a proportion, and the requirements are met, we use the sample proportion, which has (approximately) a distribution. TS 6 TRAD: Compute test stat. Sketch: Test stat; critical value(s); critical region. TS 7 TRAD: Decide. TS 6 PV: Sketch the area that is the P-value. P-value = TS 7 PV: Decide. TS 8: Statement in nontechnical terms. More: Obtain a 90% confidence interval for p. It should confirm your decision. Make a statement assessing the statistical significance of the results of this study. The P-value is a probability. Interpret it, identifying the relevant units and variable.

4 Example % of all adults in the U.S. are female. A study of state lotteries included a random digit dialing (RDD) survey. Of the surveyed adults, 248 were classified as heavy lottery players. Of these, 96 were female. Test to determine whether females are under-proportionately represented among heavy lottery players. TS 1-2-3: Hypotheses TS 4: Significance level This is a left-tailed test. Not given. When this happens you are forced to implement the P-value approach. (If you insist on an to make a decision, use 0.05.) TS 5: Because the claim is about a proportion, and the requirements are met, we use the sample proportion, which has (approximately) a Normal distribution. TS 6 PV: Compute test stat. Sketch the area that is the P-value. Note: For a left-tailed test the P-value is the area under the curve to the left of the test stat. TS 7 PV: Decide. TS 8: Statement in nontechnical terms. P-value = More: Make a statement assessing the statistical significance of the results of this study. The P-value is a probability. Interpret it, identifying the relevant units and variable.

5 Example 4 Census data suggests that in the early 1960s, 25% of those eligible for grand jury service in Alabama were black. A civil rights lawyer alleges that there has been discrimination against blacks in the actual grand jury selection (which is supposed to be done at random). Test at the 0.01 level. Of 434 jurors, 85 are black. TS 1-2-3: Hypotheses This is a -tailed test. TS 4: Significance level TS 5: Because the claim is about a proportion, and the requirements are met, we use the sample proportion, which has (approximately) a distribution. TS 6 TRAD: Compute test stat. Sketch: Test stat; critical value(s); critical region. Note: The critical region is in the left tail. TS 7 TRAD: Decide. TS 6 PV: Sketch the area that is the P-value. P-value = TS 7 PV: Decide. TS 8: Statement in nontechnical terms. More: Obtain a 98% confidence interval for p. It should confirm your decision. Make a statement assessing the statistical significance of the results of this study. The P-value is a probability. Interpret it, identifying the relevant units and variable.

6 Example 5 Some couples turn to the right, and some turn to the left, when they kiss. Do a majority turn in one direction? Test at the 5% level. Of 240 couples observed, 150 turn to the right. TS 1-2-3: Hypotheses This is a two-tailed test. TS 4: Significance level TS 5: Because the claim is about a proportion, and the requirements are met, we use the sample proportion, which has (approximately) a distribution. TS 6 TRAD: Compute test stat. Sketch: Test stat; critical value(s); critical region. Note: The critical region is split (in equal parts) over the two tails. TS 7 TRAD: Decide. TS 6 PV: Sketch the area that is the P-value. Note: For a twotailed test the P-value is twice the area under the curve to the extreme side of the test stat. P-value = TS 7 PV: Decide. TS 8: Statement in nontechnical terms. In two-tailed situations where H 0 is rejected, a statement ought to mention not only a difference but also the nature ( below or above ) of the difference. More: Obtain a 95% confidence interval for p. It should confirm your decision. Make a statement assessing the statistical significance of the results of this study. The P-value is a probability. Interpret it, identifying the relevant units and variable.

7 Solutions Example 1 H 0 : p = H 1 : p > Right tailed test. = % confidence interval: < p < Indicating we should not reject H 0. The test statistic is The critical value is We do not reject H 0. There is not sufficient sample evidence to support the claim that the new treatment curse over of all patients. (There is evidence but it is not sufficient relative to the 0.01 value protecting against falsely concluding the new cure is better, which in turn leads to the critical value that our data doesn t quite get beyond.) This difference (0.77 vs ) is not statistically significant at the 1% level. The P-value is the probability of a Z at least 2.03: P-value = = 1/47. If p = for the new treatment, then only (1 in 47) of all samples of size 100 result in a Z score of 2.03 or larger (which results from a sample proportion of 0.77 or larger). Either The null hypothesis is true and we have a fairly unlikely sample on our hands, or While the formal decision is to not reject H 0, there certainly is evidence that H 0 is false - just not enough evidence to meet the 2.326/0.01 criterion. (The standards the criterion as given by the critical value and associated significance level go together the same way that the test statistic and P-value 2.03/0.212 go together.) Example 2 H 0 : p = 0.7 H 1 : p > 0.7 Right tailed test. = pˆ = 95/126 = The test statistic is The critical value is We do not reject H 0. There is not sufficient sample evidence to support the claim that over 0.70 of all cars pass on first try. (There is a bit of evidence but it is not sufficient relative to the 0.05 value protecting against falsely concluding the rate is above 0.7.) This difference (0.754 vs. 0.7) is not statistically significant at the 5% level. 90% confidence interval: < p < Indicating we should not reject H 0. The P-value is the probability of a Z at least 1.32: P-value = = about 1/11. If p = 0.7, then (about 1 in 11) of all samples of size 126 result in a Z score of 1.32 or larger (which results from a sample proportion of or larger). Either The null hypothesis is true and we have a somewhat 6 unlikely sample on our hands, or While the formal decision is to not reject H 0, there is some evidence that H 0 is false. Not enough however to meet the 1.645/0.05 criterion. 5 This is an editorial word inserted because the P-value is rather small. I d put it in there whenever the P-value is below Another editorial word used for probabilities between about 0.05 and out of 11 is unlikely, but not highly unlikely.

8 Example 3 H 0 : p = H 1 : p < Left tailed test. = not given. pˆ = 96/248 = The test statistic is Consider = 0.10, 0.05, 0.01, and Here are the critical values: = Critical value = No matter what, the decision is to reject H 0. The P-value is the probability of a Z at most -4.14: =1/ =normsdist(-4.14) There is sufficient sample evidence to support the claim less of heavy lottery players are female. There s a lot of evidence here see the discussion of the P-value below. This difference ( vs ) is statistically significant (at any reasonable level). If p = 0.515, then (about 1 in 57586) of all samples of size 248 result in a Z score of or below (which results from a sample proportion of or less). Either The null hypothesis is true and we have a hugely 7 unlikely sample on our hands, or In this case no reasonable person would argue any way but in favor of the second choice above. Example 4 H 0 : p = 0.25 H 1 : p < 0.25 Left tailed test. = pˆ = 85/434 = The test statistic is The critical value is We reject H 0. There is sufficient sample evidence to support the claim under 0.25 of all grand jury selections are black. That is: We conclude the selection process is biased against blacks. This difference (0.196 vs. 0.25) is statistically significant at the 1% level. 98% confidence interval: < p < Indicating we should reject H 0. The P-value is the probability of a Z at most -2.60: P-value = = about 1/213. If p = 0.25, then (about 1 in 213) of all samples of size 434 result in a Z score of or below (which results from a sample proportion of or smaller). Either The null hypothesis is true and we have a quite unlikely sample on our hands, or The appropriate conclusion is that the null hypothesis is false, p < 0.25 and discrimination is taking place in some way (perhaps not consciously but it s certainly taking place). 7 Another editorial word used forreally small probabilities.

9 Example 5 H 0 : p = 0.5 H 1 : p 0.5 Two tailed test. = pˆ = 150/240 = The test statistic is The critical values are Reject H 0. There is sufficient sample evidence to support the claim that the proportion who turn right is not 0.5. (A majority turn one way.) In fact, the proportion that turns right is above 0.5. This difference (0.375 vs. 0.5) is statistically significant at the 1% level. 95% confidence interval: < p < Indicating we should reject H 0. The P-value is the probability of a Z at least 3.87 or at most -3.87: It is twice the area in the tail which is equivalent to the sum of the areas in the two tails formed by the test statistic and its opposite. Using a spreadsheet (= normsdist(-3.87): The tail area is and the P-value is twice this: = about 1 / If p = 0.5, then (about 1 in 9188) of all samples of size 240 result in a Z score beyond 3.87 in either direction (which results from a sample proportion of at most or one at least which an equal proportion turning to the left ). Either The null hypothesis is true and we have a quite unlikely sample on our hands, or The appropriate conclusion is that the null hypothesis is false, and (this is an after-the-conclusion obvious discovery) that in fact a majority turn right.

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