Question

A random sample of 328 medical doctors showed that 162 had a solo practice.

(a) Let *p* represent the proportion of all medical
doctors who have a solo practice. Find a point estimate for
*p*. (Use 3 decimal places.)

(b) Find a 95% confidence interval for *p*. (Use 3 decimal
places.)

lower limit | |

upper limit |

Give a brief explanation of the meaning of the interval. (Choose one of the below)

5% of the all confidence intervals would include the true proportion of physicians with solo practices.

5% of the confidence intervals created using this method would include the true proportion of physicians with solo practices.

95% of the confidence intervals created using this method would include the true proportion of physicians with solo practices.

95% of the all confidence intervals would include the true proportion of physicians with solo practices.

(c) As a news writer, how would you report the survey results
regarding the percentage of medical doctors in solo practice?
(Choose one of the below)

Report *p̂* along with the margin of error.

Report *p̂*.

Report the margin of error.

Report the confidence interval.

What is the margin of error based on a 95% confidence interval?
(Use 3 decimal places.) ______

Answer #1

Solution :

Given that,

n = 328

x = 162

a)

Point estimate = sample proportion = = x / n = 328 / 162 = 0.494

1 - = 1 - 0.494 = 0.506

b)

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_{/2}
= Z _{0.025} = 1.96

Margin of error = E = Z_{ / 2} * (( * (1 - )) / n)

= 1.96 * (((0.494*0.506) / 328)

= 0.054

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.494 - 0.054 < p < 0.494 + 0.054

0.440 < p < 0.548

The 95% confidence interval for the population proportion p is : (0.440 , 0.548 )

Lower limit = 0.440

Upper limit = 0.548

95% of the all confidence intervals would include the true
proportion of physicians with solo practices.

c)

Report p̂.

Margin of error = E = Z_{ / 2} * (( * (1 - )) / n)

= 1.96 * (((0.494*0.506) / 328)

= 0.054

random sample of 328 medical doctors showed that 178 had a solo
practice.
(a) Let p represent the proportion of all medical
doctors who have a solo practice. Find a point estimate for
p. (Use 3 decimal places.)
(b) Find a 98% confidence interval for p. (Use 3 decimal
places.)
lower limit
upper limit
Give a brief explanation of the meaning of the interval.
2% of the all confidence intervals would include the true
proportion of physicians with solo practices.2%...

A random sample of 320 medical doctors showed that 162 had a
solo practice.
(a) Let p represent the proportion of all medical
doctors who have a solo practice. Find a point estimate for
p. (Use 3 decimal places.)
(b) Find a 98% confidence interval for p. (Use 3 decimal
places.)
lower limit
upper limit
Give a brief explanation of the meaning of the interval.
98% of the confidence intervals created using this method would
include the true proportion of...

A random sample of 336 medical doctors showed that 160 had a
solo practice.
(a) Let p represent the proportion of all medical
doctors who have a solo practice. Find a point estimate for
p. (Use 3 decimal places.)
------------
(b) Find a 99% confidence interval for p. (Use 3 decimal
places.)
lower limit
upper limit
Give a brief explanation of the meaning of the interval.
1% of the confidence intervals created using this method would
include the...

A random sample of 324 medical doctors showed that 166 had a
solo practice.
(a) Let p represent the proportion of all medical
doctors who have a solo practice. Find a point estimate for
p. (Use 3 decimal places.)
(b) Find a 99% confidence interval for p. (Use 3 decimal
places.)
lower limit
upper limit
Give a brief explanation of the meaning of the interval.
1% of the all confidence intervals would include the true
proportion of physicians with solo...

A random sample of 322 medical doctors showed that 164 had a
solo practice.
(a) Let p represent the proportion of all medical
doctors who have a solo practice. Find a point estimate for
p. (Use 3 decimal places.)
(b) Find a 90% confidence interval for p. (Use 3 decimal
places.)
lower limit
upper limit
Give a brief explanation of the meaning of the interval.
90% of the confidence intervals created using this method would
include the true proportion of...

For this problem, carry at least four digits after the decimal
in your calculations. Answers may vary slightly due to
rounding.
In a survey of 1000 large corporations, 246 said that, given a
choice between a job candidate who smokes and an equally qualified
nonsmoker, the nonsmoker would get the job.
(a) Let p represent the proportion of all corporations
preferring a nonsmoking candidate. Find a point estimate for
p. (Round your answer to four decimal places.)
(b) Find a...

A random sample of 322 medical doctors showed that 174 had a
solo practice. (a) Let p represent the proportion of all medical
doctors who have a solo practice. Find a point estimate for p. (Use
3 decimal places.) (b) Find a 98% confidence interval for p. (Use 3
decimal places.) lower limit upper limit Give a brief explanation
of the meaning of the interval. What is the margin of error based
on a 98% confidence interval? (Use 3 decimal...

A random sample of 328 medical doctors showed that 171 had a
solo practice. Find and interpret a 95% confidence interval for the
proportion of all doctors who have a solo practice. Follow the
4-step process.
Make sure to use the 4 step process (State, Plan, Do,
Conclude)

For this problem, carry at least four digits after the decimal
in your calculations. Answers may vary slightly due to rounding. In
a marketing survey, a random sample of 1020 supermarket shoppers
revealed that 272 always stock up on an item when they find that
item at a real bargain price. (a) Let p represent the proportion of
all supermarket shoppers who always stock up on an item when they
find a real bargain. Find a point estimate for p....

For this problem, carry at least four digits after the decimal
in your calculations. Answers may vary slightly due to
rounding.
In a marketing survey, a random sample of 1000 supermarket
shoppers revealed that 268 always stock up on an item when they
find that item at a real bargain price.
(a) Let p represent the proportion of all supermarket shoppers
who always stock up on an item when they find a real bargain. Find
a point estimate for p....

ADVERTISEMENT

Get Answers For Free

Most questions answered within 1 hours.

ADVERTISEMENT

asked 5 minutes ago

asked 9 minutes ago

asked 13 minutes ago

asked 27 minutes ago

asked 35 minutes ago

asked 39 minutes ago

asked 40 minutes ago

asked 41 minutes ago

asked 41 minutes ago

asked 46 minutes ago

asked 46 minutes ago

asked 47 minutes ago