Question

QUESTION # 1

For this problem, carry at least four digits after the decimal in your calculations. Answers may vary slightly due to rounding.

In a random sample of 66 professional actors, it was found that 39 were extroverts.

(a)Let p represent the proportion of all actors who are extroverts. Find a point estimate for p. (Round your answer to four decimal places.)

(b)Find a 95% confidence interval for p. (Round your answers to two decimal places.)

lower limit

upper limit

QUESTION # 2

For this problem, carry at least four digits after the decimal in your calculations. Answers may vary slightly due to rounding.

A random sample of 5,220 permanent dwellings on an entire reservation showed that 1,669 were traditional hogans.

(a)Let p be the proportion of all permanent dwellings on the entire reservation that are traditional hogans. Find a point estimate for p.

(b)Find a 99% confidence interval for p.

Give a brief interpretation of the confidence interval.

(c)Do you think that np > 5 and nq > 5 are satisfied for this problem? Explain why this would be an important consideration.

Answer 1

1)

a)
sample proportion,= 39/66 = 0.5909

b)

sample proportion, = 0.5909
sample size, n = 66
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.5909 * (1 - 0.5909)/66) = 0.0605

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, Zc = Z(α/2) = 1.96

Margin of Error, ME = zc * SE
ME = 1.96 * 0.0605
ME = 0.1186

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.5909 - 1.96 * 0.0605 , 0.5909 + 1.96 * 0.0605)
CI = (0.47 , 0.71)

Lower limit = 0.47
upper limit = 0.71

2)

a)
sample proportion,= 1669/5220 = 0.318

b)

sample proportion, = 0.318
sample size, n = 5220
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.318 * (1 - 0.318)/5220) = 0.0064

Given CI level is 99%, hence α = 1 - 0.99 = 0.01
α/2 = 0.01/2 = 0.005, Zc = Z(α/2) = 2.58

CI(Proportion) = (0-2/34, P), P+=2 / 84,7"))

Margin of Error, ME = zc * SE
ME = 2.58 * 0.0064
ME = 0.0165

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.318 - 2.58 * 0.0064 , 0.318 + 2.58 * 0.0064)
CI = (0.3015 , 0.3345)
Therefore, based on the data provided, the 99% confidence interval for the population proportion is 0.3015 < p < 0.3345 , which indicates that we are 99% confident that the true population proportion p is contained by the interval (0.3015 , 0.3345)

c)

yes