Question

In: Statistics and Probability

The Grocery Manufacturers of America reported that 65% of consumers read the ingredients listed on a...

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).

Expert Solution

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

orchestra answered 1 year ago

The Grocery Manufactures of America reported that 76% of consumers read the ingredients listed on a...

The Grocery Manufactures of America reported that 76% of consumers read the ingredients listed on a product's label. Assume the a sample of 400 consumers is selected from the population. a. Calculate the standard error of the proportion. b. What is the probability that the sample proportion will be more than 81% of the population proportion? c. What is the probability that the sample proportion will be within ± .03 of the population proportion? d. Answer part (c) for a...

1, 33% of consumers read the ingredients listed on a product's label. For a sample of...

1, 33% of consumers read the ingredients listed on a product's label. For a sample of 265 consumers, what is the probability that between 80 and 103 of them read the ingredients listed on a product's label? 2. The hotel room rate in New York City is normally distributed with a mean of $294 per night. Assume that the standard deviation is unknown. If 23% of the New York City hotel room rates are more than $328 per night, what...

The mean cost of the ingredients of a chicken permesan meal bought at the grocery store...

The mean cost of the ingredients of a chicken permesan meal bought at the grocery store is $15. Prices from a sample of 50 restaurants for a comparable meal is shown in spreadsheet PRICES. At an alpha level of .05, can we say that eating chicken parmensan at a restaurant is cheaper? Formulate your hypotheses and solve the problem. 12.7 12.31 13.51 13.97 9.24 11.4 13.73 13.47 11.62 11.69 14.32 12.42 18.49 11.66 16.59 14.62 14.77 14.06 14.66 9.22 12.64...

. Bath salts are typically composed of the ingredients listed below. Identify each item as being...

. Bath salts are typically composed of the ingredients listed below. Identify each item as being acidic, basic, or neutral when dissolved in water. If a particular ingredient does make an acidic or basic solution, describe how this occurs. Use the results of your in-lab observations, as well as your textbook and the provided Supplemental Information on pages 65-66, to support your conclusions. NaCl NaHCO3 sodium citrate Na2SO4 MgHPO4 CaCl2 2. In Experiment SOL, you investigated the solubility of oxalic...

Prepare each solution using listed ingredients. Pay attention to the solute andsolvent. Draw the NFPA for...

Prepare each solution using listed ingredients. Pay attention to the solute andsolvent. Draw the NFPA for your final product according to the NFPA’s for each ingredient. Show your calculations. 3)Prepare a 5% light green solution with 2.5g of solute.Show Work: Weight of light green = Total Volume = 4)Prepare 50ml of 2% Congo red from a 10% Congo red solution.Show Work :Water = Volume of 10% Congo red =

The Belgian Association of Grocery Stores (BAGS) did a survey of their customers and reported that...

The Belgian Association of Grocery Stores (BAGS) did a survey of their customers and reported that 76% of them read the ingredients listed on the products labels. Assume a sample of 400 customers is selected from the population. Assume that population follows a normal distribution: a) What is the expected value of the sample proportion? b) what is the standard deviation of the sample proportion? c) what is the probability that the sample proportion will be within plus or minus...

grocery prices seem to keep going up. a market researcher was asked to interview 65 shoppers...

grocery prices seem to keep going up. a market researcher was asked to interview 65 shoppers and ask how much each spent per week to feed their family. the average amount was 166.50 with a std. deviation of $49.10. construct the 90%, 95% and 99% confidence interval for the true mean amount spent on weekly groceries.

Suppose 65% of households in America have no cats, 20% have one cat, and 15% have...

Suppose 65% of households in America have no cats, 20% have one cat, and 15% have exactly two cats.(For this question, we are assuming no one has more than two cats.) If we take a random sample of households in America, of size 20, what is the standard deviation of the corresponding sample, mean of the number of cats in each household? Round to TWO decimal places.

Carter's BBQ Restaurant produces BBQ sauce for resale at grocery stores throughout North America. The company...

Carter's BBQ Restaurant produces BBQ sauce for resale at grocery stores throughout North America. The company is currently in the process of establishing a master budget on a quarterly basis for this coming fiscal year, which ends December 31. Prior year quarterly sales were as follows (1 unit = 1 batch): First quarter 67,200 units Second quarter 80,640 units Third quarter 100,800 units Fourth quarter 87,360 units Unit sales are expected to increase 25%, and each unit is expected to...

You own a grocery store in the city of St. Louis. As you read the paper...

You own a grocery store in the city of St. Louis. As you read the paper this morning you discover that a proposal to fund a new football stadium in St. Louis includes raising the sales tax in the city of St. Louis by making it 0.5% higher for all sales within the city limits. Compose a letter to the editor of your newspaper to explain your point of view on this proposal. Make sure your letter addresses the specific...