Question

The following 5 questions are based on this information. Census indicates that the proportion of adult men in the United States is 49% (p=0.49).

We plan to take a random sample of 300 U.S. adults.

1) The sampling distribution of p-bar is

Select one:

a. normal because np≥5   and n(1−p)≥5

b. normal because np≤5 and n(1−p)≤5

c.  not normal because the sample size is too small

d.  not normal because  np≤5

2) The standard error (SE) of   p-bar is

Select one:

a. 0.001

b. 0.025

c. 0.028

d. 0.006

3) What is the probability that a random sample of 300 U.S. adults will provide a sample proportion (p bar) that is within 0.04 of the population proportion (p)?

Select one:

a. 23%

b. 83.42%

c. 43%

d. 77%

4) What is the probability that a random sample of 300 U.S. adults will provide a sample proportion (p bar) that is within 0.07 of the population proportion (p)?

Select one:

a. 98.47%

b. 16%

c. 84%

d. 99.968%

5) Say, you took a random sample of 300 U.S. adults, and found out that the sample proportion (p bar) for this sample to be 0.38 (or 38%). In light of the result from the PRECEDING question,

Select one:

a.  This is not considered a strange finding because a sample proportion of 38% is quite likely

b. This is considered a very strange finding because a sample proportion of 38% should be extremely rare

Answer 1

1)

a. normal because np≥5   and n(1−p)≥5

2)

std error of proportion=σp=√(p*(1-p)/n)=

0.028

3)

probability =P(0.45<X<0.53)=P((0.45-0.49)/0.029)<Z<(0.53-0.49)/0.029)=P(-1.39<Z<1.39)=0.9171-0.0829=0.8342~83.42%

4)

probability =P(0.42<X<0.56)=P((0.42-0.49)/0.029)<Z<(0.56-0.49)/0.029)=P(-2.43<Z<2.43)=0.9924-0.0076=0.9847~ 98.47%

5)

b. This is considered a very strange finding because a sample proportion of 38% should be extremely rare