Question

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 5805 physicians in Colorado showed that 3035 provided at least some charity care (i.e., treated poor people at no cost).

(a) Let p represent the proportion of all Colorado physicians who provide some charity care. Find a point estimate for p. (Round your answer to four decimal places.)

(b) Find a 99% confidence interval for p. (Round your answers to three decimal places.)

lower limit

upper limit

Part 2

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 64 professional actors, it was found that 40 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

Answer 1

(1)

(a)

n = Sample Size = 5805

p = Sample Proportion = 3035/5805 =0.5228

so,

point estimate = 0.5228

(b)

q =1 - p = 0.4772

SE =

= 0.01

From Table, critical values of Z = 2.576

Confidence Interval:

0.5228 (2.576 X 0.006555)

= 0.5228 0.0169

= ( 0.5059, 0.5397)

So,

Confidence Interval:

0.5059 <P < 0.5397

So

Lower Limit = 0.5059

Upper limit = 0.5397

(2)

(1)

(a)

n = Sample Size = 64

p = Sample Proportion = 40/64 =0.6250

so,

point estimate = 0.6250

(b)

q =1 - p = 0.3750

SE =

= 0.05

From Table, critical values of Z = 1.96

Confidence Interval:

0.6250 (1.96 X 0.0605)

= 0.6250 0.1186

= ( 0.5064, 0.7436)

So,

Confidence Interval:

0.5064 <P < 0.7436

So

Lower Limit = 0.5064

Upper limit = 0.7436