Chapter 9 Inferences from Two Samples 9-2 Inferences About Two Proportions 1. There are two requirements for using the methods of this section, and each of them is violated. (1) The samples should be two sample random samples that are independent. These samples are convenience samples, not simple random samples. These samples are likely not independent. Since she surveyed her friends, she may well have males and females that are dating each other (or least that associate with each other) – and people tend to associate with those that have similar behaviors. (2) The number of successes for each sample should be at least 5, and the number of failures for each sample should be at least 5. This is not true for the males, for which x=4. NOTE: This is the same requirement from previous chapters for using the normal distribution to approximate the binomial that required np≥5 and nq≥5. Using ˆp=x/n to estimate p and ˆq= 1- x/n = (n-x)/n to estimate q, nˆp≥5 nˆq≥5 n[x/n]≥5 n[(n-x)/n] ≥5 x≥5 (n-x) ≥5 These inequalities state that the number of successes must be greater than 5, and the number of failures must be greater than 5. 2. We have 95% confidence that the limits of -0.0518 and 0.0194 contain the true difference between the population proportions of subjects who experience headaches. Repeating the trials many times would result in confidence limits that would include the true difference between the population proportions 95% of the time. Since the interval includes the value 0, there is no significant difference between the two populations proportions. 3. In this context, 1ˆp= 15/1583 = 0.00948 2ˆp= 8/157 = 0.05096 p = (15+8)/(1583+157) = 23/1740 = 0.01322 p1denotes the rue proportion of all Zocor users who experience headaches p2denotes the true proportion of all placebo users who experience headaches 4. No. The P-value method and the traditional method will always agree, but it is possible for the confidence interval approach to lead to a different conclusion. The P-value and traditional methods used a standard deviation for the sampling distribution of 1ˆp - 2ˆpbased on the hypothesized difference of zero between the population proportions. With no specific hypothesis as part of its procedure, the confidence interval method uses a standard deviation for the sampling distribution of 1ˆp - 2ˆpbased on solely on the individual sample proportions. Specifically, the P-value method and the traditional method use 12ˆˆp -p12pqpqσ=+nnwhile the confidence interval method uses 121122ˆˆp -p12ˆ ˆˆ ˆp qp qσ=+nn.
Inferences About Two Proportions SECTION 9-2 281 5. x = (0.158)(8834) = 1395.772 ≈1396 NOTE: Any 1392 ≤x ≤1400 rounds to x/8834 = 15.8%. 6. x = (0.124)(129) = 15.996 ≈16 7. 1ˆp= x1/n1= 13/36 = 0.361 1ˆp - 2ˆp= 0.361 – 0.519 = -0.157 2ˆp= x1/n1= 14/27 = 0.519 a. p = (x1+x2)/(n1+n2) = (13+14)/(36+27) = 27/63 = 0.429 b.