Question

For each of the following problems, you must:

a) Define the population and the parameter of interest;

b) Identify the given and unknown information and write them in correct notations,

c) Verify if the sampling distribution is approximately normal by checking the conditions np > 5 and nq > 5),

d) Compute the probability.

e) Interpret the answer.

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1) The article “Should Pregnant Women Move? Linking Risks for Birth Defects with Proximity to Toxic Waste Sites” (Chance (1992): 40-45) reported that in a large study carried out in the state of New York, approximately 30% of the population lived within 1 mile of a hazardous waste site. If an SRS of 400 pregnant women is selected, how likely is it that the sample proportion will be within 5% of the true population proportion? Would this probability be larger or smaller if we selected an SRS of size 500?

2) The article “Thrillers” (Newsweek, Apr. 22, 1985) states, “Surveys tell us that more than half of America’s college graduates are avid readers of mystery novels.” Assume the true proportion is exactly 0.5. What is the probability that an SRS of 225 college graduates would give a sample proportion greater than 0.6?

3) Suppose that a particular candidate for public office is in fact favored by 48% of all registered voters in a sizable metropolitan district. A polling organization takes an SRS of 500 voters and will use the sample proportion to estimate the population parameter. What is the probability that the sample proportion will be greater than 0.5, causing the polling organization to incorrectly predict the results of the upcoming election?

Answer 1

1) n = 400

P = 0.3

np = 400 * 0.3 = 120

nq = 400 * (1 - 0.3) = 280

So the sampling distribution is approximately normal, because np > 5 and nq > 5.

= p = 0.3

= sqrt(p(1 - p)/n)

= sqrt(0.3 * 0.7/400)

= 0.0229

P(0.25 < < 0.35)

= P((0.25 - )/ < ( - )/ < (0.35 - ​​​​​​)/)

= P((0.25 - 0.3)/0.0229 < Z < (0.35 - 0.3)/0.0229)

= P(-2.18 < Z < 2.18)

= P(Z < 2.183) - P(Z < -2.183)

= 0.9855 - 0.0145

= 0.9710

2) n = 225

p = 0.5

np = 225 * 0.5 = 112.5

nq = 225 * 0.5 = 112.5

So the sampling distribution is approximately normal because np > 5 and nq > 5.

= 0.5

= sqrt(p(1 - p)/n)

= sqrt(0.5 * 0.5/225)

= 0.0333

P( > 0.6)

= (( - )/ > (0.6 - ​​​​​​)/)

= P(Z > (0.6 - 0.5)/0.0333)

= P(Z > 3.003)

= 1 - P(Z < 3.003)

= 1 - 0.9987

= 0.0013

3) n = 500

p = 0.48

np = 500 * 0.48 = 240

nq = 500 * (1 - 0.48) = 260

So the sampling distribution is approximately normal because np > 5 and nq > 5.

= 0.48

= sqrt(p(1 - p)/n)

= sqrt(0.48 * 0.52/500)

= 0.0223

P( > 0.5)

= (( - )/ > (0.5 - ​​​​​​)/)

= P(Z > (0.5 - 0.48)/0.0223)

= P(Z > 0.897)

= 1 - P(Z < 0.897)

= 1 - 0.8151

= 0.1849