Question

Thirty-nine 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 population proportion of St. Paulites who drink bottled water more than once a week is 0.39, the same as the overall proportion of Americans who drink bottled water more than once a week. Use z-table.

a. Suppose you select a sample of 540 St.Paulites. Show the sampling distribution of p (to 4 decimals).

E(p)=

Op=

b. Based upon a sample of 540 St. Paulites, what is the probability that the sample proportion will be within 0.09 of the population proportion (to 4 decimals).

probability =

c. Suppose you select a sample of 250 St.Paulites. Show the sampling distribution of p (to 4 decimals).

E(p) =

Op=  

d. Based upon a smaller sample of only 250 St. Paulites, what is the probability that the sample proportion will be within 0.09 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)?

Reduced by =

Everything in Bold is what needs to be answered.

Answer 1

the PDF of normal distribution is = 1/σ * √2π * e ^ -(x-u)^2/ 2σ^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
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(a) when sample of 540 St.Paulites selected
proportion ( p ) = 0.39
standard Deviation ( sd )= sqrt(PQ/n) = sqrt(0.39*0.61/540)
=0.021
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(b)
the probability that the sample proportion will be within 0.09 of the population proportion is ( 0.39-0.09 < X < 0.39 + 0.09)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 0.3) = (0.3-0.39)/0.021
= -0.09/0.021 = -4.2857
= P ( Z <-4.2857) From Standard Normal Table
= 0.00001
P(X < 0.48) = (0.48-0.39)/0.021
= 0.09/0.021 = 4.2857
= P ( Z <4.2857) From Standard Normal Table
= 0.99999
P(0.3 < X < 0.48) = 0.99999-0.00001 = 1
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(c)
when sample of 250 St.Paulites selected
proportion ( p ) = 0.39
standard deviation ( sd )= sqrt(PQ/n) = sqrt(0.39*0.61/250) =0.0308
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(d)
the probability that the sample proportion will be within 0.09 of the population proportion is ( 0.39-0.09 < X < 0.39 + 0.09)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 0.3) = (0.3-0.39)/0.0308
= -0.09/0.0308 = -2.9221
= P ( Z <-2.9221) From Standard Normal Table
= 0.00174
P(X < 0.48) = (0.48-0.39)/0.0308
= 0.09/0.0308 = 2.9221
= P ( Z <2.9221) From Standard Normal Table
= 0.99826
P(0.3 < X < 0.48) = 0.99826-0.00174 = 0.9965
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(e)
by taking the larger sample in parts (a) and (b) rather than the smaller sample in parts (c) and (d) we gain is = 1 - .9965 = 0.0035 = 0.35%