8.3-Testing a Claim about a Proportion: Objectives: 1. Test claims about a proportion using the P-value Method. 2. Test claims about a proportion using the Traditional Method 3. Test claims about a proportion using Confidence Intervals
Overview: This section presents complete procedures for testing a hypothesis (or claim) made about a population proportion. This section uses the components introduced in the previous section for the P-value method, the traditional method or the use of confidence intervals.
Basic Methods of Testing Claims about a Population Proportion: The objective is to test a claim about a population proportion using a formal method of hypothesis testing.
Notation: n = sample size or number of trials x = sample proportion pˆ = n p = population proportion q=1-p
Requirements: 1. The sample observations are a simple random sample. 2. The conditions for a binomial distribution are satisfied. 3. The conditions np ≥ 5 and nq ≥ 5 are both satisfied, so the binomial distribution of sample proportions can be approximated by a normal distribution with µ = np and s = npq. Note: p is the assumed proportion not the sample proportion.
Test Statistic for Testing a Claim about a Proportion: z=
pˆ − p pq n
P-values: Use the standard normal distribution Table A-2 Critical Values: Use the standard normal distribution: Table A-2
Caution: 1. Do not confuse P-value with p. The P-value is the probability of getting a test statistic at least as extreme as the one representing the sample data and p is the population proportion. 2. Sometimes the value of pˆ is given directly and sometimes it must be determined from the information:
Comprehensive Hypothesis Test: The following procedure was introduced in the last set of notes. It combines the basic components of hypothesis testing into a comprehensive procedure.
General Procedure for P-Value Method: 1. Identify the specific claim or hypothesis to be tested, and put it in symbolic form. 2. Give the symbolic form that must be true when the original claim is false. 3. Of the two symbolic expressions obtained so far, let the alternative hypothesis be the one not containing equality. Let the null hypothesis be the symbolic expression that the parameter equals the fixed value being considered. 4. Select the significance level α based on the seriousness of type 1 error. Make α small if the consequences of rejecting a true H 0 are severe. The values of 0.05 and 0.01 are very common. 5. Identify the statistic that is relevant to this test and determine its sampling distribution (normal or t-distribution). 6. Find the test statistic and find the P-value. Draw a graph and show the test statistic and the Pvalue. 7. Reject H 0 if the P-value is less than or equal to the significance level α . Fail to reject H 0 if the P-value is greater than α . 8. Restate this previous decision in simple non-technical terms and address the original claim.
Traditional Method: (First 5 steps are the same as the P-value method) 6. Find the test statistic, critical values and the critical region. Draw a graph and include the test statistic, critical values, and critical region. 7. Reject H 0 if the test statistic is in the critical region. Fail to reject H 0 if the test statistic is not in the critical region. 8. Restate this previous decision in simple non-technical terms and address the original claim.
Confidence Interval Method: A confidence interval estimate of a population parameter contains the likely values of that parameter. We should therefore reject a claim that the population parameter has a value that is not included in the confidence interval.
In some cases, a conclusion based on a confidence interval may be different from a conclusion based on a hypothesis test. Use a confidence interval to estimate a population proportion, but use the P-value or traditional method for testing a claim about a proportion.
Example: A study was conducted in which 57 out of 104 pregnant women correctly guessed the sex of their babies. Use these sample data to test the claim that the success rate of such guesses is no different from the 50% success rate expected with random chance guesses. Use a 0.05 significance level. Solution: Requirements are satisfied: simple random sample; fixed number of trials (104) with two categories (guess correctly or do not); np = (104)(0.5) = 52 ≥ 5 and nq = (104)(0.5) = 52 ≥ 5 Step 1: Original claim is that the success rate is no different from 50%: p = 0.50 Step 2: Opposite of original claim is p ≠ 0.50 Step 3: Because p ≠ 0.50 does not contain equality so it is H1. Therefore, H0: p = 0.50 is the null hypothesis and original claim and H1: p ≠ 0.50 is the alternative hypothesis. Step 4: Significance level is α = 0.05 Step 5: The sample involves a proportion so the relevant statistic is the sample proportion pˆ . Step 6: calculate z:
57 − 0.5 ˆp − p z= = 104 = 0.98 pq (0.5)(0.5) n 104 This is a two-tailed test; P-value is twice the area to the right of test statistic. Table A-2: z = 0.98 has an area of 0.8365 to its left, so area to the right is 1 – 0.8365 = 0.1635, doubles yields 0.3270. Step 7: Because the P-value of 0.3270 is greater than the significance level of 0.05, fail to reject the null hypothesis. Step 8: Here is the correct conclusion: There is not sufficient evidence to warrant rejection of the claim that women who guess the sex of their babies have a success rate equal to 50%.
Example: A supplier of digital memory cards claims that no more than 1% of the cards are defective. In a random sample of 600 memory cards, it is found that 3% are defective, but the supplier claims that this is only a sample fluctuation. At the 0.01 level of significance, test the supplier's claim that no more than 1% are defective. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim. Solution: Requirements are satisfied: simple random sample; fixed number of trials (600) with two categories (defective or not); np = (600)(0.01) = 6 ≥ 5 and nq = (600)(0.99) = 594 ≥ 5 Step 1: Original claim is that no more than 1% of the memory cards are defective. p ≤ 0.01 . Step 2: Opposite of original claim is p > 0.01 Step 3: Because p > 0.01 does not contain equality it is H1. Therefore, H0: p ≤ 0.01 is the null hypothesis and original claim and H1: p > 0.01 is the alternative hypothesis. Step 4: Significance level is α = 0.01 Step 5: The sample involves a proportion so the relevant statistic is the sample proportion pˆ . Step 6: calculate z:
z=
pˆ − p pq n
=
0.03 − 0.01 (0.01)(0.99) 600
=
0.02 = 22.22 0.0009
This is a right-tailed test; P-value is the area to the right of test statistic. Table A-2: z = 22.22 has an area of 0.9999 to its left, so area to the right is 1 – 0.9999 = 0.0001. Step 7: Because the P-value of 0.0001 is less than the significance level of 0.01, reject the null hypothesis. Step 8: Here is the correct conclusion: There is sufficient evidence to reject the claim that no more than 1% of the cards are defective.
Example: A poll of 1068 adult Americans reveals that 48% of the voters surveyed prefer the Democratic candidate for the presidency. At the 0.05 level of significance, test the claim that at least half of all voters prefer the Democrat. Identify the null hypothesis, alternative hypothesis, test statistic, Pvalue, conclusion about the null hypothesis, and final conclusion that addresses the original claim. Solution: Requirements are satisfied: simple random sample; fixed number of trials (1068) with two categories (democrat or not); np = (1068)(0.5) = 534 ≥ 5 and nq = (1068)(0.5) = 534 ≥ 5 Step 1: Original claim is that at least half of all voters prefer the Democrat p ≥ 0.5 . Step 2: Opposite of original claim is p < 0.5 Step 3: Because p < 0.5 does not contain equality so it is H1. Therefore, H0: p ≥ 0.5 is the null hypothesis and original claim and H1: p < 0.5 is the alternative hypothesis. Step 4: Significance level is α = 0.05 Step 5: The sample involves a proportion so the relevant statistic is the sample proportion pˆ . Step 6: calculate z:
z=
pˆ − p pq n
=
0.48 − 0.5 (0.5)(0.5) 1068
=
− 0.02 = −1.31 0.01529
This is a left-tailed test; P-value is the area to the left of test statistic. Table A-2: z = -1.31 has an area of 0.0951 to its left. Step 7: Because the P-value of 0.0951 is greater than the significance level of 0.05, fail to reject the null hypothesis. Step 8: Here is the correct conclusion: There is not sufficient evidence to warrant rejection of the claim that at least half of all voters prefer the Democrat.
Example: An article in a journal reports that 34% of American fathers take no responsibility for child care. A researcher claims that the figure is higher for fathers in the town of Littleton. A random sample of 234 fathers from Littleton yielded 96 who did not help with child care. Test the researcher's claim at the 0.05 significance level. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim. Solution: Requirements are satisfied: simple random sample; fixed number of trials (234) with two categories (guess correctly or do not); np = (234)(0.34) = 80 ≥ 5 and nq = (234)(0.66) = 154 ≥ 5 Step 1: Original claim is that more than 34% of fathers in Littleton take no responsibility for child care. p > 0.34 Step 2: Opposite of original claim is p ≤ 0.34 Step 3: Because p > 0.34 does not contain equality, it is H1. Therefore, H0: p ≤ 0.34 is the null hypothesis and original claim and H1: p > 0.34 is the alternative hypothesis. Step 4: Significance level is α = 0.05 Step 5: The sample involves a proportion so the relevant statistic is the sample proportion pˆ . Step 6: calculate z:
z=
pˆ − p pq n
=
96 − 0.34 0.0703 234 = = 2.28 (0.34)(0.66) 0.0309 234
This is a right-tailed test; P-value is the area to the right of test statistic. Table A-2: z = 2.28 has an area of 0.9887 to its left, so area to the right is 1 – 0.9887 = 0.0113. Step 7: Because the P-value of 0.0113 is less than the significance level of 0.05, reject the null hypothesis. Step 8: Here is the correct conclusion: The sample data support the claim that more than 34% of fathers in Littleton take no responsibility for child care.
Caution: When testing claims about a population proportion, the traditional method and the P-value method are equivalent and will yield the same result since they use the same standard deviation based on the claimed proportion p. However, the confidence interval uses an estimated standard deviation based upon the sample proportion p. Consequently, it is possible that the traditional and P-value methods may yield a different conclusion than the confidence interval method. A good strategy is to use a confidence interval to estimate a population proportion, but use the P-value or traditional method for testing a claim about the proportion.
