AP Statistics Testing Hypothesis About Proportions Chapter 20

Objectives Hypothesis Null hypothesis Alternative hypothesis Two-sided alternative One-sided alternative P-value One-proportion z-test

Significance Testing Used to investigate preconceived assumptions about some condition in the population. Usually this condition can be expressed as a mean of some characteristic or as a proportion of some characteristic of interest. Sample data are selected and either the sample mean or proportion is calculated in order to determine if this value could reasonably be assumed to exist w/in the hypothesized population.

Logic of Tests of Significance In statistical testing, we want to show whether a certain claim about the value of a parameter is reasonable or not. For the test, we determine the criteria under which we will conclude that the assumption is unreasonable, take an appropriate sample and calculate the relevant statistic from the data, and then compare the results to our criteria.

General Procedure for One-Proportion z-Test P H A N T O M S P arameter – State the population parameter of interest. H ypothesis – State the Null Hypothesis and Alternative Hypothesis. A ssumptions – Verify the conditions for the test. N ame – Name the hypothesis test to be used. T est Statistic – Calculate the test statistic from the sample data. O btain P-value – Use the test statistic to calculate the p-value. M ake Decision – Make the decision to reject or fail to reject the null hypothesis. S tate Conclusion in Context – Using the p-value and your decision above, state your conclusion in the context of the problem.

Hypothesis A statement of a condition which is assumed to exist in a population and is tested using the results from a randomly selected sample.

Null and Alternative Hypothesis Null Hypothesis (H0) – the hypothesis being tested. Usually a “no change” or “no difference” statement about a parameter (mean or proportion) of the distribution. Example: p = p0 (an equal sign should appear in the null hypothesis). Generally, it is the null hypothesis that the researcher is hoping to reject in favor of a proposed alternative hypothesis.

Alternative Hypothesis (Ha or H1) – the alternative to the null hypothesis. Often it is this hypothesis that the researcher hopes to prove true. Three choices possible for the alternative hypothesis. If the primary concern is deciding whether a population proportion, p, is different from a specified value p0, the alternative hypothesis should be p≠p0. Express as: Ha:p≠p0 A hypothesis test of this form is called a two-tailed or two-sided test.

If the primary concern is deciding whether a population proportion, p, is less than a specified value p0, the alternative hypothesis should be p<p0. Express as: Ha:p<p0 A hypothesis test of this form is called a one-sided or one-tailed (left-tailed) test. If the primary concern is deciding whether a population proportion, p, is greater than a specified value p0, the alternative hypothesis should be p>p0. Express as: Ha:p>p0 A hypothesis test of this form is called a one-sided or one-tailed (right-tailed) test. A hypothesis test is called a one-tailed test if it is either left-tailed or right-tailed, that is, if it is not two-tailed.

Illustration

1 - Choosing the Null and Alternative Hypotheses A large city’s Department of Motor Vehicle’s claimed that 80% of candidates pass driving tests, but a newspaper’s survey of 90 randomly selected local teens who had taken the test found only 61 (68%) who passed. Does this finding suggest that the passing rate for teenagers is lower than the DMV reported? Determine the null hypothesis for the hypothesis test. Determine the alternative hypothesis for the hypothesis test. Classify the hypothesis test as two-tailed, left-tailed, or right-tailed.

1 - Solution The null hypothesis is: The passing rate for teenagers is 80%, as the DMV claimed. H0: p = .80 The alternative hypothesis is: The passing rate for teenagers is less than the 80% claimed by the DMV. Ha: p < .80 This hypothesis test is (single-tail) left-tailed.

2 - Choosing the Null and Alternative Hypotheses Advances in medical care such as prenatal ultrasound examination now make it possible to determine a child’s sex early in pregnancy. There is a fear that in some cultures some parents may use this technology to select the sex of their children. A study for India reports that, in 1993, in one hospital, 56.9% of the live births that year were boys. It’s a medical fact that male babies are slightly more common than female babies. The study’s authors report a baseline for this region of 51.7% male live births. Is there evidence that the proportion of male births has changed? Determine the null hypothesis for the hypothesis test. Determine the alternative hypothesis for the hypothesis test. Classify the hypothesis test as two-tailed, left-tailed, or right-tailed.

2 - Solution The null hypothesis is: The proportion of male births has not changed and is still equal to the baseline of 51.7%. H0: p = .517 The alternative hypothesis is: The proportion of male births has changed and is no longer equal to the baseline of 51.7%. Ha: p ≠ .517 This hypothesis test is two-tailed.

Test Statistic A sample statistic or value based on the sample data. The test statistic is used as a basis for deciding whether the null hypothesis should be rejected or not. Is a z value for the sample

Critical Value or P-Value? The decision to reject or fail to reject the null hypothesis can be made by comparing the test statistic to a critical value (based on a confidence level) or by comparing a p-value (based on the test statistic) to a significance level.

P-Values A P-value is a conditional probability. The probability that we obtain the value of the test statistic that we observed or a value that is more extreme in the direction of Ha, given that H0 is true. It is the probability of the observed test statistic given that the null hypothesis is true. P-value = (observed statistic value[or more extreme]|H0) The P-value is not the probability that the null hypothesis is true.

P-Values The smaller the P-value, the more strongly we are inclined to reject H0. If the P-value is very small, it is very unlikely that a value as extreme as the observed value of the test statistic would be the outcome if H0 were true. To obtain the P-value of a hypothesis test, we assume that the null hypothesis is true and compute the probability of observing a value of the test statistic as extreme or more extreme than that observed. This is the area of the tail (relative to the test statistic observed) under the standard normal curve.

z0 is the observed value of the test statistic z Illustration: z0 is the observed value of the test statistic z

Example: Calculating P-value Test Statistic z = 1.71 (two-tailed) P-value = 2•P(z>1.71) P-value = 2•normalcdf(1.71,100) P-value = .08727

Example: Calculating P-value Test Statistic z = 2.85 (right-tailed) P-value = P(z>2.85) P-value = normalcdf(2.85,100) P-value = .002186

Example: Calculating P-value Test Statistic z = -.88 (left-tailed) P-value = (z<-.88) P-value = normalcdf(-100,-.88) P-value = .189

P-Values When the data are consistent with the model from the null hypothesis, the P-value is high and we are unable to reject the null hypothesis. In that case, we have to “retain” the null hypothesis we started with. We can’t claim to have proved it; instead we “fail to reject the null hypothesis” when the data are consistent with the null hypothesis model and in line with what we would expect from natural sampling variability. If the P-value is low enough, we’ll “reject the null hypothesis,” since what we observed would be very unlikely were the null model true.

P-Values and Decisions: What to Tell About a Hypothesis Test How small should the P-value be in order for you to reject the null hypothesis? It turns out that our decision criterion is context-dependent. When we’re screening for a disease and want to be sure we treat all those who are sick, we may be willing to reject the null hypothesis of no disease with a fairly large P-value (0.10). A longstanding hypothesis, believed by many to be true, needs stronger evidence (and a correspondingly small P-value) to reject it. Another factor in choosing a P-value is the importance of the issue being tested.

P-Values and Decisions Your conclusion about any null hypothesis should be accompanied by the P-value of the test. If possible, it should also include a confidence interval for the parameter of interest. Don’t just declare the null hypothesis rejected or not rejected. Report the P-value to show the strength of the evidence against the hypothesis. This will let each reader decide whether or not to reject the null hypothesis.

A Trial as a Hypothesis Test – The Logic of a Significance Test Think about the logic of jury trials: To prove someone is guilty, we start by assuming they are innocent. We retain that hypothesis until the facts make it unlikely beyond a reasonable doubt. Then, and only then, we reject the hypothesis of innocence and declare the person guilty.

A Trial as a Hypothesis Test The same logic used in jury trials is used in statistical tests of hypotheses: We begin by assuming that a hypothesis is true. Next we consider whether the data are consistent with the hypothesis. If they are, all we can do is retain the hypothesis we started with. If they are not, then like a jury, we ask whether they are unlikely beyond a reasonable doubt.

What to Do with an “Innocent” Defendant If the evidence is not strong enough to reject the presumption of innocent, the jury returns with a verdict of “not guilty.” The jury does not say that the defendant is innocent. All it says is that there is not enough evidence to convict, to reject innocence. The defendant may, in fact, be innocent, but the jury has no way to be sure.

What to Do with an “Innocent” Defendant Said statistically, we will fail to reject the null hypothesis. We never declare the null hypothesis to be true, because we simply do not know whether it’s true or not. Sometimes in this case we say that the null hypothesis has been retained.

What to Do with an “Innocent” Defendant In a trial, the burden of proof is on the prosecution. In a hypothesis test, the burden of proof is on the unusual claim. The null hypothesis is the ordinary state of affairs, so it’s the alternative to the null hypothesis that we consider unusual (and for which we must marshal evidence).

One-Proportion z-Test The conditions for the one-proportion z-test are the same as for the one proportion z-interval. We test the hypothesis H0: p = p0 using the test statistic where When the conditions are met and the null hypothesis is true, this statistic follows the standard Normal model, so we can use that model to obtain a P-value.

Assumptions and Conditions for One-Proportion z-Test The same as for the sampling distribution of sample proportions and the one-proportion z-interval. The assumptions and the corresponding conditions must be checked before conducting a Hypothesis Test for a proportion: Independence Assumption: We first need to Think about whether the Independence Assumption is plausible. It’s not one you can check by looking at the data. Instead, we check two conditions to decide whether independence is reasonable.

Assumptions and Conditions for One-Proportion z-Test Randomization Condition: Were the data sampled at random or generated from a properly randomized experiment? Proper randomization can help ensure independence. 10% Condition: Is the sample size no more than 10% of the population? Sample Size Assumption: The sample needs to be large enough for us to be able to use the CLT. Success/Failure Condition: We must expect at least 10 “successes” and at least 10 “failures.”

Procedure: Hypothesis Test for Population Proportions P H A N T O M S Parameter Hypothesis Assumptions Name the Test Test Statistic Z test statistic Obtain p-value Make a decision State conclusion in context

Example: A large city’s Department of Motor Vehicle’s claimed that 80% of candidates pass driving tests, but a newspaper’s survey of 90 randomly selected local teens who had taken the test found only 61 (68%) who passed. Does this finding suggest that the passing rate for teenagers is lower than the DMV reported?

Solution P H A N T O M S Parameter: p0: 80% of candidates pass driving tests. Hypothesis: Null hypothesis: H0: p=.80 H0: The passing rate for teenagers is 80%, as the DMV claimed. Alternative hypothesis: Ha: p<.80 Ha: The passing rate for teenagers is less than the 80% claimed by the DMV.

Solution Assumptions: Randomization Condition: The 90 teens were a random sample. 10% Condition: 90 is less than 10% of teenagers taking driving tests in a large city. Success/Failure Condition: np0=90(.80)=72>10 and nq0=90(.20)=18>10. Name the Test: One proportion z-test

Solution Test statistic

Solution Obtain p-value p-value = .00192 Make Decision The p-value is small enough to reject the null hypothesis in favor of the alternative. Conclusion in context Because the p-value of .00192 is very low, I reject the null hypothesis. This data provides strong evidence that the passing rate for teenagers taking the driving test is less the 80%.

Another Example: In a given year, 13.55% of employed people in the United States reported belonging to a union. Officials from a large city contacted a random sample of 2000 city workers and 240 claimed union membership. Is there sufficient evidence to conclude that the proportion of works in this city who are union members is different from the national rate?

Solution P H A N T O M S Parameter: p0: 13.55% of employed people in the United States belong to a union. Hypothesis: Null hypothesis: H0: p=.135 H0: The proportion of union members in this city is equal to the national rate of .135. Alternative hypothesis: Ha: p≠.135 Ha: The proportion of union members in this city is different from the national rate of .135.

Solution Assumptions: Randomization Condition: random sample is stated. 10% Condition: 10n≤N, 20,000 is less than all the city workers if the city is large. Success/Failure Condition: np0=2000(.135)=270>10 n(1-p0)=2000(.865)=1730>10 Name the Test: One proportion z-test

Solution Test Statisic p0=.135 z = -1.97

Solution Obtain p-value Make Decision Conclusion in context p-value = 2P(z > 1.97) Because it is a two-sided test, the p-value = 2 [normalcdf(1.97,100)] p-value = .0488 Make Decision The p-value is small enough to reject the null hypothesis in favor of the alternative. Conclusion in context With a p-value of .0488 (< .05), there is sufficient evidence to conclude that, in this city, the proportion of workers who are union members is different from the national value.

Your Turn: A 1996 report from the U.S. Consumer Product Safety Commission Claimed that at least 90% of all American homes have at least one smoke detector. A city’s fire department has been running a public safety campaign about smoke detectors consisting of posters, billboards, and ads on radio and TV and in the newspaper. The city wonders if this concerted effort has raised the local level above the 90% national rate. Building inspectors visit 400 randomly selected homes and find that 376 have smoke detectors. Is this strong evidence that the local rate is higher than the national rate?

Using the TI-84 STAT/TESTS/1-PropZTest Input p0: x: the number selected n: the sample size Select type of test: ≠,<,> Calculate

Solve using the TI-84 Input p0: .8 x: 61 n: 90 A large city’s Department of Motor Vehicle’s claimed that 80% of candidates pass driving tests, but a newspaper’s survey of 90 randomly selected local teens who had taken the test found only 61 (68%) who passed. Does this finding suggest that the passing rate for teenagers is lower than the DMV reported? Input p0: .8 x: 61 n: 90 Select type of test: < Calculate

Solution: prop<.8 z=-2.898754522 p=.0018733072 𝑝 =.6777777778 n=90 Because the P-value of .00187 is very low, I reject the null hypothesis. This data provides strong evidence that the passing rate for teenagers taking the driving test is less the 80%.

Your Turn: Solve using the TI-84 In a given year, 13.55% of employed people in the United States reported belonging to a union. Officials from a large city contacted a random sample of 2000 city workers and 240 claimed union membership. Is there sufficient evidence to conclude that the proportion of works in this city who are union members is different from the national rate?