Question

Thirty-seven percent of all Americans drink bottled water more than once a week (Natural resources Defense Council, December 4, 2015). Suppose you have been hired by the Natural Resources Defence Council to investigate bottled water consumption in St. Paul. You plan to select a sample of St. Paulites to estimate the proportion who drink bottled water more than once a week. Assume the popluation proportion of St. Paulites who drink bottled water more than once a week is 0.37, the same as the overall proportion of Americans who drink bottled water more than once a week. Use z-table.

b. Based upon a sample of 540 St. Paulites, what is the probability that the sample proportion will be within 0.06 of the population proportion (to 4 decimals).

probability = ________

  

d. Based upon a smaller sample of only 230 St. Paulites, what is the probability that the sample proportion will be within 0.06 of the population proportion (to 4 decimals).

probability = ___________

e. As measured by the increase in probability, how much do you gain in precision by taking the larger sample in parts (a) and (b) rather than the smaller sample in parts (c) and (d)?

- Select your answer -Reduced byIncreased = ____________

- Select your answer -Have gain in precision by increasing the sample.

Have gain in precision by decreasing the sample.

Answer 1

Answer:

Given,

n = 540

p^ = 0.37

Standard deviation = sqrt(p^(1-p^)/n)

= sqrt(0.37(1-0.37)/540)

= 0.0208

b)

To give P(-0.06 < pbar - p < 0.06)

= P(-0.06/0.0208 < z < 0.06/0.0208)

= P(-2.8846 < z < 2.8846)

= P(z < 2.88) - P(z < -2.88)

= 0.9980 - 0.0019

= 0.9961

d)

Here n = 230

Standard deviation = sqrt(p^(1-p^)/n)

= sqrt(0.37(1-0.37)/230)

= 0.0318

P(- 0.06 < p^-p < 0.06)

= P(-0.06/0.0318 < z < 0.06/0.0318)

= P(-1.89 < z < 1.89)

= P(z < 1.89) - P(z < - 1.89)

= 0.970621 - 0.029379

= 0.9412

e)

Here it is reduced by 0.0549 .

Hence there will be gain in precision by increasing the sample size.