Chapter 9: TESTS OF HYPOTHESES FOR A SINGLE SAMPLE Part 3: Hypothesis tests on a population proportion Statistical significance vs. Practical significance Sections 9-5 and 9-1.6
• Recall, for large n we have . p(1−p) ˆ P ∼ N (p, n )
where Pˆ is our point estimator for the population proportion p. The sampling distribution for Pˆ is approximately normal, and this approximation relies on np > 5 and n(1 − p) > 5, and n large.
1
• We can use this normal approximation to perform a hypothesis test on p. H0 : p = p0 H1 : p 6= p0 (2-sided test) • Test statistic: Pˆ − p0 Z0 = q
p0(1−p0) n
under H0 true, Z0 ∼ N (0, 1)
• So, under the null being true, we know the behavior of Pˆ .
2
• Example: Voters in favor of an issue A random sample of 500 registered voters in Phoenix is asked if they favor the use of oxygenated fuels year-round to reduce air pollution and 332 respond positively. Does the data present strong evidence that at least 60% of the voters in Phoenix are in favor of the use of the fuels? Perform a hypothesis test at the α = 0.05 significance level. ANS:
3
• Example: Belief in ghosts A newspaper reported that the proportion of the population that believe in ghosts is 0.44. You think that this number is something else. You ask 64 people and 36 of them say that they believe in ghosts. Conduct the appropriate hypothesis test using a significance level of 0.05. ANS:
4
Connection between hypothesis tests and confidence intervals Part of section 9-1
• There is a close relationship between a 100(1-α)% confidence interval for µ and a 2-sided hypothesis test for µ done at the α significance level.
• Example: Volume of juice in cans (previous example) Hypothesis test performed at α = 0.05 level. H0 : µ = 12 H1 : µ 6= 12 √ Test statistic: z0 = 11.98−12 = −1.05 0.19/ 100
p-value = 2 × P (Z ≤ −1.05) = 0.2925 5
⇒ Fail to reject at 0.05 level. µ = 12 could’ve reasonably generated the sample data.
Connection to 95% confidence interval for µ √
x¯ ± z0.025(s/ n) ⇒ [11.943, 12.017] 95% CI contains 12, µ = 12 is a plausible population mean. ———————————————————— Both the 95% CI and the hypothesis test performed at the α = 0.05 give the same conclusion, µ = 12 is reasonable.
6
• This connection exists in general. • Example: Belief in ghosts (previous example) Hypothesis test performed at α = 0.05 level. H0 : p = 0.44 H1 : p 6= 0.44 q Test statistic: z0 = 0.56−0.44 = 1.97 0.44(.56) 64
p-value = 2 × P (Z ≥ 1.97) = 0.0482 ⇒ Reject at 0.05 level, and we say there is strong evidence against p = 0.44. Connection to 95% confidence interval for p q pˆ(1−ˆ p) pˆ ± z0.025 ⇒ [0.4410, 0.6840] n 7
95% CI does NOT contain 0.44, p = 0.44 is NOT a plausible population proportion based on this CI. ———————————————————– Both the 95% CI and the hypothesis test performed at the α = 0.05 give the same conclusion, p = 0.44 is rejected.
8
Statistical significance vs. Practical significance Part of section 9-1
Consider a drug company that has been working on a new headache relief product. They would like to put the new drug into production if it relieves pain much quicker. The present drug relieves pain in 15 minutes, on average. H0 : µ = 15 H1 : µ < 15 {one-sided test} Suppose time to relief follows a normal distribution with a known standard deviation σ = 4 minutes. 2 4 ¯ Under H0, X ∼ N (15, n )
Suppose a random sample of n = 25 is taken. ¯ under H0 true Below is the distribution of X and n = 25. 9
0.5 0.4 0.0
0.1
0.2
0.3
Distribution of X−bar under Hnull true Rejection Region for X−bar shown in red (left of 13.684) one−sided test
11
12
13
14
X−bar
15
16
17
For α = 0.05, we will reject H0 in favor of H1 if x¯ < 13.684. ———————————————————– Suppose the new drug is better (H0 is false), and it’s average time to relief is 14 minutes. 2 4 ¯ Under H1 true and µ = 14, X ∼ N (14, n )
We have set our critical value at x¯ = 13.684 to control the type I error rate at the 0.05 level (we only reject if x¯ is below this value). 10
18
What is the probability that we reject under this scenario (i.e. what is the power of the test)? ¯ under H1 true and µ = 14 The distribution of X is shown below (a shift left compared to H0).
0.0
0.1
0.2
0.3
0.4
0.5
Distribution of X−bar under H1 true (mu=14)
11
12
13
14
X−bar
15
16
17
The probability of rejection is the area in green (includes red overlay), ¯ ≤ 13.684 |H1 true ) = 0.3464 P (X With this sample size, we have a 35% chance of detecting the improvement of the new drug. 11
18
If the random sample had a sample mean x¯ = 13.8 minutes, we do not reject H0, and we accept µ = 15. We would not have strong statistical evidence that µ < 15, and that the new drug is better. ————————————————————– What if we have a sample size n = with x¯ = 13.8. What is the result of the hypothesis test? ¯ under H0 true and under The distribution of X H1 true are both shown below for n = 100. Both are more narrow with the larger n. Distribution of X−bar under both scenarios
11
12
13
14
12
X−bar
15
16
17
18
We now have a critical value of x¯ = 14.342 to control the type I error rate at the 0.05 level (we only reject if x¯ is below this value). The probability of rejection given H1 is true is the area in green (includes red overlay), ¯ ≤ 14.342 |H1 true ) = 0.8037 P (X With this larger sample size, we have an 80% chance of detecting the improvement of the new drug.
13
Though the two scenarios haven’t changed, we now have a larger n, reducing the variability in ¯ around its mean, and we have much more X power of detection. Distribution of X−bar under both scenarios
11
12
13
14
X−bar
15
16
17
For n = 100 and and an observed x¯ = 13.8, we do reject H0 in favor of H1. By increasing n, we were able to get statistically significant evidence that H0 is false in favor of H1 : µ < 15. BUT IS THIS A PRACTICAL SIGNIFICANCE DIFFERENCE? 14
18
Is a 1 minute decrease in time to relief going to sell more medicine? Will they recoup their costs of changing all their manufacturing process? Be careful because you can make a difference in values be statistically significant with a large enough n, but it has to be a meaningful difference for the problem at hand to change the status quo.
15
