1. 1 Section 8.3 Testing a claim about a Proportion Objective For a population with proportion p, use a sample (with a sample proportion) to test a claim about the proportion. Testing a proportion uses the standard normal distribution (z-distribution)

2. 2 Notation

3. 3 (1) The sample used is a a simple random sample (i.e. selected at random, no biases) (2) Satisfies conditions for a Binomial distribution (3) n p 0 ≥ 5 and n q 0 ≥ 5 Requirements Note: p 0 is the assumed proportion, not the sample proportion Note: 2 and 3 satisfy conditions for the normal approximation to the binomial distribution

4. 4 Test Statistic Denoted z (as in z-score) since the test uses the z-distribution.

5. 5 If the test statistic falls within the critical region, reject H 0. If the test statistic does not fall within the critical region, fail to reject H 0 (i.e. accept H 0 ). Traditional method:

6. 6 Types of Hypothesis Tests: Two-tailed, Left-tailed, Right-tailed The tails in a distribution are the extreme regions where values of the test statistic agree with the alternative hypothesis

7. 7 Left-tailed Test “<” H 0 : p = 0.5 H 1 : p < 0.5  significance level Area =  -z  (Negative)

8. 8 Right-tailed Test “>” H 0 : p = 0.5 H 1 : p > 0.5  significance level Area =  zz (Positive)

9. 9 Two-tailed Test “≠” H 0 : p = 0.5 H 1 : p ≠ 0.5  significance level z  Area =  -z 

10. 10 The XSORT method of gender selection is believed to increases the likelihood of birthing a girl. 14 couples used the XSORT method and resulted in the birth of 13 girls and 1 boy. Using a 0.05 significance level, test the claim that the XSORT method increases the birth rate of girls. (Assume the normal birthrate of girls is 0.5) What we know: p 0 = 0.5 n = 14 x = 13 p = 0.9286 Claim: p > 0.5 using α = 0.05 Example 1 n p 0 = 14*0.5 = 7 n q 0 = 14*0.5 = 7 Since n p 0 > 5 and n q 0 > 5, we can perform a hypothesis test.

11. 11 H 0 : p = 0.5 H 1 : p > 0.5 Example 1 Right-tailed What we know: p 0 = 0.5 n = 14 x = 13 p = 0.9286 Claim: p > 0.5 using α = 0.01 z in critical region z = 3.207 z α = 1.645 Test statistic: Critical value: Initial Conclusion: Since z is in the critical region, reject H 0 Final Conclusion: We Accept the claim that the XSORT method increases the birth rate of girls

12. 12 P-Value The P-value is the probability of getting a value of the test statistic that is at least as extreme as the one representing the sample data, assuming that the null hypothesis is true. z Test statistic z α Critical value z zαzα P-value = P(Z > z) p-value (area) Example

13. 13 P-Value Critical region in the left tail: Critical region in the right tail: Critical region in two tails: P-value = area to the left of the test statistic P-value = area to the right of the test statistic P-value = twice the area in the tail beyond the test statistic

14. 14 P-Value method: If the P is low, the null must go. If the P is high, the null will fly. If P-value  , reject H 0. If P-value > , fail to reject H 0.

15. 15 Caution Don’t confuse a P-value with a proportion p. Know this distinction: P-value = probability of getting a test statistic at least as extreme as the one representing sample data p = population proportion

16. 16 Calculating P-value for a Proportion Stat → Proportions → One sample → with summary

17. 17 Calculating P-value for a Proportion Enter the number of successes (x) and the number of observations (n)

18. 18 Calculating P-value for a Proportion Enter the Null proportion (p 0 ) and select the alternative hypothesis (≠, ) Then hit Calculate

19. 19 Calculating P-value for a Proportion The resulting table shows both the test statistic (z) and the P-value Test statistic P-value P-value = 0.0007

20. 20 Using P-value Initial Conclusion: Since p-value < α (α = 0.05), reject H 0 Final Conclusion: We Accept the claim that the XSORT method increases the birth rate of girls P-value = 0.0007 Stat → Proportions→ One sample → With summary Null: proportion= Alternative Number of successes: Number of observations: H 0 : p = 0.5 H 1 : p > 0.5 Example 1 What we know: p 0 = 0.5 n = 14 x = 13 p = 0.9286 Claim: p > 0.5 using α = 0.01 13 14 0.5 > ● Hypothesis Test

21. 21 Do we prove a claim? A statistical test cannot definitely prove a hypothesis or a claim. Our conclusion can be only stated like this: The available evidence is not strong enough to warrant rejection of a hypothesis or a claim We can say we are 95% confident it holds. “The only definite is that there are no definites” -Unknown

22. 22 Mendel’s Genetics Experiments When Gregor Mendel conducted his famous hybridization experiments with peas, one such experiment resulted in 580 offspring peas, with 26.2% of them having yellow pods. According to Mendel’s theory, ¼ of the offspring peas should have yellow pods. Use a 0.05 significance level to test the claim that the proportion of peas with yellow pods is equal to ¼. What we know: p 0 = 0.25 n = 580 p = 0.262 Claim: p = 0.25 using α = 0.05 Example 2 n p 0 = 580*0.25 = 145 n q 0 = 580*0.75 = 435 Since n p 0 > 5 and n q 0 > 5, we can perform a hypothesis test. Problem 32, pg 424

23. 23 H 0 : p = 0.25 H 1 : p ≠ 0.25 Example 2 Two-tailed z not in critical region z = 0.667 z α = -1.960 Test statistic: Critical value: Initial Conclusion: Since z is not in the critical region, accept H 0 Final Conclusion: We Accept the claim that the proportion of peas with yellow pods is equal to ¼ What we know: p 0 = 0.25 n = 580 p = 0.262 Claim: p = 0.25 using α = 0.05 z α = 1.960

24. 24 Using P-value Example 2 Initial Conclusion: Since P-value > α, accept H 0 Final Conclusion: We Accept the claim that the proportion of peas with yellow pods is equal to ¼ H 0 : p = 0.25 H 1 : p ≠ 0.25 What we know: p 0 = 0.25 n = 580 p = 0.262 Claim: p = 0.25 using α = 0.05 x = np = 580*0.262 ≈ 152 P-value = 0.5021 Stat → Proportions→ One sample → With summary Null: proportion= Alternative Number of successes: Number of observations: 152 580 0.25 ≠ ● Hypothesis Test