Question

The Grocery Manufacturers of America reported that 61% of consumers read the ingredients listed on a product's label. Assume the population proportion is p = 0.61 and a sample of 380 consumers is selected from the population a. Show the sampling distribution of the sample proportion p, where p is the proportion of the sampled consumers who read the ingredients listed on a product's label E(F) = (to 2 decimals) (to 4 decimals) b, what is the probability that the sample proportion will be within ±0.03 of the population proportion (to 4 decimals). c. Answer part (b) for a sample of 590 (to 4 decimals)

Answer 1

The population proportion is p=0.61 and the sample size is n= 380. a. The mean and the standard deviation of the sampling distribution of the sample proportion D is calculated as follows: E(B)=p=0.61 p(1- p)_ 0.61(1-0.61) – 0.025 Opvn V 380 Hence, the mean of the sampling distribution of the sample proportion is 0.61 and the standard deviation of the sampling distribution of the sample proportion is 0.025.

+0.03 of the population The probability that the sample proportion will be within proportion is calculated as follows: P(p-pl<0.03)=P(-0.03<p-p<0.03) = P(ē- p<0.03) – P(-p<-0.03) 0.03 =pl_- p P(1-P) 0 .03 0.025 P-p p(1-p) 0.025 = P(Z <1.2) - P(Z <-1.2) = 0.8849 – 0.1151 (From Z-table P(Z <1.2) = 0.8849 and P(Z <-1.2)=0.1151) = 0.7698 Hence, the probability that the sample proportion will be within +0.03 of the population proportion is 0.7698 The sample size is n= 590.

+0.03 of the population The probability that the sample proportion will be within proportion is calculated as follows: P(7 - p <0.03)=P(-0.03 < 5 - p<0.03) = P(P-p<0.03)-P(-p<-0.03) p-p < < PIZ 0.02 0.03 P-P 0.03 | P(1–p) 10.61(1-0.61) P(1- p) 0.61(1 – 0.61) n v 590 ) V 590 pha 0.03) ole 0.03 = P Z 0.02 = P(Z <1.5)- P(Z <-1.5) = 0.9332 -0.0668 (From Z-table P(Z <1.5)= 0.9332 and P(Z <-1.5) = 0.0668) = 0.8664 Hence, the probability that the sample proportion will be within +0.03 of the population proportion is 0.8664.