Question

The Grocery Manufacturers of America reported that 65% of consumers read the ingredients listed on a product's label. Assume the population proportion is p=0.65 and a sample of 500 consumers is selected from the population.

(a) Show the sampling distribution of the sample proportion (p-bar) where, (p-bar) is the proportion of the sampled consumers who read the ingredients listed on a product's label. (1) E(p)= (xxxx) (2 decimals) (2) std dev of (p-bar) = (xxxx)

(b) What is the probability that the sample proportion will be within .03 of the population proportion (to 4 decimals).

(c) . Answer part (b) for a sample of 600 (to 4 decimals).

Answer 1

population proportion of consumers reading the ingredients listed on a product's label is, p=0.65

sample size, n = 500

a) From this

sample proportion, E(p) = population proportion = p = 0.65

E(p) = 0.65

sample variance = p*(1-p)/n = 0.65*(1-0.65)/500 = 0.65*0.35/500 = 0.2275/500 = 0.000455

sample standard deviation = sqrt( sample variance ) = sqrt( 0.000455 ) = 0.0213

std dev of (p-bar) = 0.0213

b) Let P denotes the sample proportion

Probability that the sample proportion will be within .03 of the population proportion

P[ 0.65 - 0.03 < P < 0.65 + 0.03 ] = P[ 0.62 < P < 0.68 ] = P[ ( 0.62 - 0.65 )/0.0213 < ( P - 0.65 )/0.0213 < ( 0.68 - 0.65 )/0.0213 ] = P[ -1.41 < Z < 1.41 ] = 2*P[ 0 < Z < 1.41 ] = 2*0.4207 = 0.8414

Probability that the sample proportion will be within .03 of the population proportion = 0.8414

c) If sample size increases to 600 sample proportion will be same but the variance will change.

sample variance = p*(1-p)/n = 0.65*(1-0.65)/600 = 0.65*0.35/600 = 0.2275/600 = 0.000379

sample standard deviation = sqrt( sample variance ) = sqrt( 0.000379 ) = 0.0195

std dev of (p-bar) = 0.0195

Probability that the sample proportion will be within .03 of the population proportion

P[ 0.65 - 0.03 < P < 0.65 + 0.03 ] = P[ 0.62 < P < 0.68 ] = P[ ( 0.62 - 0.65 )/0.0195 < ( P - 0.65 )/0.0195 < ( 0.68 - 0.65 )/0.0195 ] = P[ -1.54 < Z < 1.54 ] = 2*P[ 0 < Z < 1.54 ] = 2*0.4382 = 0.8764

Probability that the sample proportion will be within .03 of the population proportion = 0.8764