Week 10 Chapter 16. Confidence Intervals for Proportions

Point and Interval Estimation A point estimate, for example, estimated mean or estimated proportion, is a single number that is the best value for the parameter (e.g., population mean, or population proportion). Example: In 2009, a Canadian General Social Survey on Criminal Victimization reported that 0.076 of the Canadians who were surveyed think their neighbourhood has higher than average crime. An interval estimate is an interval of numbers around the point estimate, within which the parameter value is believed to fall. Same Example above: An interval estimate predicts that the population proportion, of all Canadians, who think their neighbourhood has higher than average crime falls between 0.071 and 0.081. This interval predicts that the point estimate of 0.076 falls within margin of error of 0.005 of the true population proportion: 0.076 ± 0.005. That is, the sample proportion and the true proportion should be less than 0.005 apart.

Sampling distribution of sample proportion 𝑝 Recall: Draw a simple random sample of size n from a large population proportion 𝑝 of success. 𝑝 falls between 0 and 1. Let X be the count of success in the sample and 𝑝 = 𝑋 𝑛 the sample proportion of successes. 𝑝 falls between 0 and 1. When n is large, the sampling distribution of these statistics are approximately normal: X is approx. Normal (𝑛𝑝, 𝑛𝑝(1−𝑝) ) 𝑝 is approx. Normal (𝑝, 𝑝(1−𝑝) 𝑛 ) Note that sample proportion 𝑝 is a sample mean (we count the number of successes and divide by n), therefore, Central Limit Theorem applies. That is sample proportion, 𝑝 is approximately normal with mean 𝑝 and standard error of 𝜎 𝑝 = 𝑝(1−𝑝) 𝑛 . Note in practice the value for 𝑝 is unknown, thus, the value for the standard error is unknown. In this case we use the information from the sample and estimate the standard error (se). 𝜎 𝑝 =𝑠𝑒= 𝑝 (1− 𝑝 ) 𝑛 *Criteria (guide): Sampling distribution is approx. normal when 𝑛𝑝≥10 𝑎𝑛𝑑 𝑛𝑝(1−𝑝)≥10. Since sampling distribution of sample proportion 𝑝 is approximately normal we can apply the Empirical Rule (since we have a symmetric bell-shaped) and use the Z-table (since we have an approximately normal distribution).

Confidence Interval (CI) CI is interval estimates that contains the parameter with a certain degree of confidence (correct inference). CI is Point Estimate ± Margin of Error. Confidence level is the probability that this method, CI, produces an interval that contains the parameter. This number is close to 1 (never 1 as it would mean to have all possible values of point estimate, for example sample proportion, for a characteristic of a population (e.g., population proportion)); thus, this number is usually, 0.90, 0.95, or 0.99.

Confidence Interval is Long-Run Proportion Correct A confidence interval constructed from any particular sample either does or does not contain the population parameter, for example true proportion 𝑝. If we repeatedly selected random samples of that size and each time constructed a 95% confidence interval, then in the long run about 95% of the intervals (19 out of 20 times) would contain the population parameter, for example true proportion 𝑝. On average, only about 5% (about 1 out of 20 times) does a 95% confidence interval fail to contain the population parameter. Note that different samples have different 𝑝 ’s, and give different confidence intervals. So our confidence in procedure rather than an individual interval.

Large-Sample CI for a Proportion According to the Empirical Rule, about 95% of the normal distribution falls within 2 standard deviation of the mean; more precisely using the Z-table, 1.96 standard deviation. Recall, Z for 95% interval: Left over is 5% = 0.05 Half that is 2.5% = 0.025. 0.025 proportion is above Z = 1.96 So, with probability 0.95, 𝑝 falls within 1.96 𝜎 𝑝 of 𝑝: (𝑝-1.96 𝜎 𝑝 , 𝑝 +1.96 𝜎 𝑝 )

Large-Sample CI for a Proportion Once sample is selected, if 𝑝 does fall within 1.96 𝜎 𝑝 units of 𝑝, then the interval from 𝑝 -1.96 𝜎 𝑝 to 𝑝 +1.96 𝜎 𝑝 contains 𝑝. So, with probability 0.95, a 𝑝 value occurs such that the interval 𝑝 ±1.96 𝜎 𝑝 contains the population parameter 𝑝. On the other hand, the probability is 0.05 [1- confidence level(0.95) = 0.05] that 𝑝 does not fall within 1.96 𝜎 𝑝 of 𝑝. In that case, the interval from 𝑝 -1.96 𝜎 𝑝 to 𝑝 +1.96 𝜎 𝑝 does not contains 𝑝. 0.05 is called the error probability and it is denoted by 𝛼 (the Greek letter alpha). Recall in practice the value for 𝑝 is unknown, thus, the value for the standard error is unknown. In this case we use the information from the sample and estimate the standard error (se). 𝜎 𝑝 =𝑠𝑒= 𝑝 (1− 𝑝 ) 𝑛 The general form for the confidence interval for a population proportion 𝑝 is: 𝑝 ± z(𝑠𝑒) where z(se) is called the margin of error.

Z-score for the Usual Confidence Levels Confidence Level (%): (1- 𝜶) x 100 𝜶 𝜶/2 Z-score 90% 0.10 0.05 1.65 (or 1.64, or 1.645) 95% 0.025 1.96 99% 0.01 0.005 2.58 (or 2.57, or 2.575)

Example: 99% CI for a Proportion In 2009, a Canadian General Social Survey on Criminal Victimization reported that of the 19,422 Canadians surveyed, 1476 of them think their neighbourhood has higher than average crime. Construct the 99% CI for the true proportion of Canadians who think their neighbourhood has higher than average crime. Construct the 99% CI for the true proportion of Canadians who do not think their neighbourhood has higher than average crime.

Example: 99% CI for a Proportion Construct the 99% CI for the true proportion of Canadians who think their neighbourhood has higher than average crime. 99% CI for true proportion 𝑝: 𝑝 ± z(𝑠𝑒) = 0.076 ± 2.58 (0.0019) = 0.076 ± 0.0049 = (0.071, 0.081) 𝑝 = 1476 19422 = 0.076 Z for 99% interval: Left over is 1% = 0.01 Half that is 0.5% = 0.005. Cannot find exactly 0.005 proportion. 0.0049 proportion is above Z = 2.58 𝑠𝑒= 𝑝 (1− 𝑝 ) 𝑛 = 0.076 (1−0.076) 19422 = 0.0019 Interpretation: We are 99% Confident that the true proportion of Canadians who think their neighbourhood has higher than average crime is between 0.071 and 0.081.

Example: 99% CI for a Proportion Construct the 99% CI for the true proportion of Canadians who do not think their neighbourhood has higher than average crime. Let 1−𝑝 denote proportion of Canadians who do not think their neighbourhood has higher than average crime. 99% CI for true proportion 1−𝑝: (1- 𝑝 ) ± z(𝑠𝑒) 1- 𝑝 = 1- 0.076 = 0.924 Z = 2.58 𝑠𝑒= 𝑝 (1− 𝑝 ) 𝑛 = 1−0.076 0.076 19422 = 0.0019 99% CI for true proportion 1−𝑝: (1- 𝑝 ) ± z(𝑠𝑒) = 0.924 ± 2.58 (0.0019) = 0.924 ± 0.0049 = (0.919, 0.929)* *Verify: Subtract each endpoint of the 99% CI for true proportion 𝑝: (0.071, 0.081) from 1.0 1- 0.071 = 0.929 1- 0.081 = 0.919

The width of a CI increases as the Confidence Level increases c. Construct the 90% CI for the true proportion 𝑝: 𝑝 ± z(𝑠𝑒) = 0.076 ± 1.645 (0.0019) = 0.076 ± 0.003125 = (0.073, 0.079) d. Construct the 95% CI for the true proportion 𝑝: 𝑝 ± z(𝑠𝑒) = 0.076 ± 1.96 (0.0019) = 0.076 ± 0.0037 = (0.072, 0.080) And recall a. 99% CI for true proportion 𝑝: 𝑝 ± z(𝑠𝑒) = 0.076 ± 2.58 (0.0019) = 0.076 ± 0.0049 = (0.071, 0.081)

The width of a CI decreases as the sample size increases Suppose in a random sample of size (n) 4855, the proportion of Canadians who think their neighbourhood has higher than average crime is 0.076. 99% CI for true proportion 𝑝: 𝑝 ± z(𝑠𝑒) = 0.076 ± 2.58 (0.0038) = 0.076 ± 0.0098 = (0.0662, 0.0858) This interval is bigger and it is in this case twice as wide as the one below with n = 19422. 𝑝 = 0.076 Z for 99% interval: Z = 2.58 𝑠𝑒= 𝑝 (1− 𝑝 ) 𝑛 = 0.076 (1−0.076) 4855 = 0.0038 recall for n = 19422 with 𝑝 = 0.076, 99% CI for true proportion 𝑝: 𝑝 ± z(𝑠𝑒) = 0.076 ± 2.58 (0.0019) = 0.076 ± 0.0049 = (0.071, 0.081) The width of this interval is smaller and it is in this case half the size of the one above with n = 4855.

Use a CI for Making an Inference Suppose it is claimed that 10% of Canadians think their neighbourhood has higher than average crime. In 2009, a Canadian General Social Survey on Criminal Victimization reported that of the 19,422 Canadians surveyed, 1476 of them think their neighbourhood has higher than average crime. Is there evidence to suggest that percentage of Canadians who think their neighbourhood has higher than average crime is different from 10%? 10% = 0.10 is the claimed (true) proportion. 95% CI for true proportion 𝑝: (0.072, 0.080) does not contain 0.10. Thus, we can conclude that the true proportion of Canadians who think their neighbourhood has higher than average crime is different from 0.10 (it is between 0.072 and 0.080). Moreover, both values in this CI are less than 0.10. This suggests that the percentage of Canadians who think their neighbourhood has higher than average crime is less than 0.10.