Hypothesis Tests for a Population Proportion

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1 Hypothesis Tests for a Population Proportion MATH 130, Elements of Statistics I J. Robert Buchanan Department of Mathematics Fall 2015

2 Review: Steps of Hypothesis Testing 1. A statement is made regarding the nature of the population (usually about µ or p). 2. Evidence (sample data) is collected to test the statement. 3. The data are analyzed to assess the plausibility of the statement.

3 Background Hypothesis testing about the population proportion is carried out very similarly to the familiar method for hypothesis testing involving the population mean. Assumptions: 1. Simple random sample of size n 0.05N is collected. 2. If p 0 is the assumed value of the population proportion, then np 0 (1 p 0 ) The test statistic will be calculated as z = ˆp p 0 p 0 (1 p 0 ) n.

4 Methods There are three equivalent methods for conducting the hypothesis test: 1. Classical approach, 2. P-Value approach, 3. Confidence interval approach.

5 Classical Approach 1. State the null and alternative hypotheses. H 0 : p = p 0 H 1 : p, <, > p 0 2. Select a level of significance α. 3. Calculate the test statistic: z = ˆp p 0 p 0 (1 p 0 ) n 4. If the test statistic falls in the critical region, reject H State the conclusion..

6 Critical Region (1 of 2) Test Two-tailed CV ±z α/2 CR

7 Critical Region (2 of 2) Test Left-tailed Right-tailed CV z α z α CR

8 Remarks If the sample mean is too many standard deviations away from the mean stated in the null hypothesis, we reject the null hypothesis. When the results observed in the sample are unlikely to have occurred under the assumption of the null hypothesis, we say the result is statistically significant. The classical approach is robust in the sense that small departures from the assumption of a normally distributed population do not usually affect the result of the test.

9 Classical Approach (1 of 3) An insurance company states that 90% of its claims are settled within 30 days. A consumer group selected a simple random sample of 75 of the company s claims to test this statement. The consumer group found that 55 of the claims were settled within 30 days. At the 0.05 significance level, test the company s claim that 90% of its claims are settled within 30 days.

10 Classical Approach (1 of 3) An insurance company states that 90% of its claims are settled within 30 days. A consumer group selected a simple random sample of 75 of the company s claims to test this statement. The consumer group found that 55 of the claims were settled within 30 days. At the 0.05 significance level, test the company s claim that 90% of its claims are settled within 30 days. Remark: the phrase 90% of claims implies 90% or more. The consumer group s hypothesis is that less than 90% of claims are settled in 30 days.

11 Claasical Approach (2 of 3) H 0 : p = 0.90 H 1 : p < 0.90 (left-tailed test) α = 0.05, z α = z 0.05 = Test statistic: z = 55/ (1 0.90) 75 = 4.811

12 Example (3 of 3) TS: z Decision: reject H 0. Conclusion: the sample data warrants rejection of the claim that 90% of the company s insurance claims are settled within 30 days (at the α = 0.05 significance level).

13 P-Value Approach (1 of 3) A politician claims that she will receive at least 60% of the votes in an upcoming election. The results of a simple random sample of 100 voters showed that 58 of those sampled will vote for her. Test the politician s claim at the 0.05 level of significance.

14 P-Value Approach (1 of 3) A politician claims that she will receive at least 60% of the votes in an upcoming election. The results of a simple random sample of 100 voters showed that 58 of those sampled will vote for her. Test the politician s claim at the 0.05 level of significance. Remark: the politician is stating that she will receive 60% or more of the votes. The alternative is that she will receive less than 60% of the votes.

15 P-Value Approach (2 of 3) H 0 : p = 0.60 H 1 : p < 0.60 (left-tailed test) α = 0.05, z α = z 0.05 = Test statistic: z = 58/ (1 0.60) 100 = 0.41

16 P-Value Approach (3 of 3) P-value: Decision: do not reject H 0. P(Z < 0.41) = = α Conclusion: the sample data do not warrant rejection of the claim that the politician will receive 60% of the vote (at the α = 0.05 level of significance with P-value = ).

17 Classical Approach (1 of 3) The full-time student body of a college is 50% men and 50% women. Suppose an introductory chemistry class contains 30 men and 20 women. Does this sample provide sufficient evidence at the 0.05 significance level to reject the hypothesis that the proportions of male and female students who take this course are the same as in the general student body?

18 Classical Approach (2 of 3) H 0 : p = 0.50 H 1 : p 0.50 (two-tailed test) α = 0.05, α/2 = 0.025, ±z α/2 = ±1.96 Test statistic: z = 30/ (1 0.50) 50 = 1.41

19 Classical Approach (3 of 3) TS: z Decision: fail to reject H 0. Conclusion: at the α = 0.05 significance level, the sample data does not warrant rejection of the claim that the proportions of male and female students in the introductory chemistry class is the same as in the general student body.

20 P-Value Approach (1 of 3) The popularity of personal watercraft (also known as jet skis) continues to increase, despite the apparent danger associated with their use. A sample of 54 watercraft accidents reported to the Nebraska Game and Parks Commission in 1997 revealed that 85% of them involved personal watercraft even though only 8% of the motorized boats registered in the state are personal watercraft. Suppose the national average proportion of watercraft accidents in 1997 involving personal watercraft was 78%. Does the watercraft accident rate in Nebraska exceed the rate in the nation? Use the 0.01 level of significance.

21 P-Value Approach (2 of 3) H 0 : p = 0.78 H 1 : p > 0.78 (right-tailed test) α = 0.01, z α = 2.33 Test statistic: z = (1 0.78) 54 = 1.24

22 P-Value Approach (3 of 3) P-value: P(Z > 1.24) = 1 P(Z < 1.24) Decision: do not reject H 0. = = = α Conclusion: at the α = 0.01 significance level, the sample data do not support that claim that the watercraft accident rate in Nebraska exceeds the rate in the nation (P-value = ).

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