Question

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

You need to convert your interval to z-scores .25 and .35

These are proportions. To convert them (or standardizethem) you need the formula z =

p is the proportion you wish to standardize, π is givento be .3

and n is given to be 400.

The denominator of the formula is the Standard Error(SE)

Start with .25

Numerator: .25 - .30 = -0.05

Denominator: √[.30(1-.30) / 400] =√[.30(.70) / 400] = 0.0229

so z = -0.05 / 0.0229 = -2.182

Now .35

Numerator: .35 - .30 = 0.05

Denominator = 0.0229

z = 0.05 / 0.0229 = 2.182

Now use the computer or a Z-table to find the area between-2.182 and 2.182

So now the P(-2.182 < z < 2.182) = P(.25 < p <.35) = 0.9709