Question

In: Statistics and Probability

A random sample of 356 medical doctors showed that 175 had a solo practice. As a...

A random sample of 356 medical doctors showed that 175 had a solo practice. As a news writer, how would you report the survey results regarding the percentage of medical doctors in solo practice? What is the margin of error based on a 90% confidence interval?

Group of answer choices

A recent study shows that about 49% of medical doctors have a solo practice with a margin of error of 2.2 percentage points.

A recent study shows that about 49% of medical doctors have a solo practice with a margin of error of 4.4 percentage points.

A recent study shows that about 51% of medical doctors have a solo practice with a margin of error of 8.7 percentage points.

A recent study shows that about 49% of medical doctors have a solo practice with a margin of error of 8.7 percentage points.

A recent study shows that about 51% of medical doctors have a solo practice with a margin of error of 4.4 percentage points.

Flag this Question

Question 21 pts

A random sample of medical files is used to estimate the proportion p of all people who have blood type B. If you have no preliminary estimate for p, how many medical files should you include in a random sample in order to be 99% sure that the point estimate  will be within a distance of 0.04 from p?

Group of answer choices

17

33

2081

4,161

1,041

Flag this Question

Question 31 pts

Suppose a certain species bird has an average weight of  = 3.85 grams. Based on previous studies, we can assume that the weights of these birds have a normal distribution with σ = 0.31 grams. Find the sample size necessary for a 98% confidence level with a maximal error of estimate E = 0.09 for the mean weights of the hummingbirds.

Group of answer choices

183

23

65

517

9

Flag this Question

Question 41 pts

Suppose twenty-two communities have an average of  = 139.6 reported cases of larceny per year. Assume that σ is known to be 45.4 cases per year. Find a 90%, 95%, and 98% confidence interval for the population mean annual number of reported larceny cases in such communities. Compare the margins of error. As the confidence level increase, do the margins of error increase?

Group of answer choices

The 90% confidence level has a margin of error of 74.7; the 95% confidence level has a margin of error of 89.0; and the 98% confidence level has a margin of error of 105.8. As the confidence level increases, the margins of error increase.

The 90% confidence level has a margin of error of 3.4; the 95% confidence level has a margin of error of 4.0; and the 98% confidence level has a margin of error of 4.8. As the confidence level increases, the margins of error increase.

The 90% confidence level has a margin of error of 105.8; the 95% confidence level has a margin of error of 89.0; and the 98% confidence level has a margin of error of 74.7. As the confidence level increases, the margins of error decrease.

The 90% confidence level has a margin of error of 22.6; the 95% confidence level has a margin of error of 19.0; and the 98% confidence level has a margin of error of 15.9. As the confidence level increases, the margins of error decrease.

The 90% confidence level has a margin of error of 15.9; the 95% confidence level has a margin of error of 19.0; and the 98% confidence level has a margin of error of 22.6. As the confidence level increases, the margins of error increase.

Flag this Question

Question 51 pts

Suppose a certain species bird has an average weight of  = 3.10 grams. Based on previous studies, we can assume that the weights of these birds have a normal distribution with σ = 0.21 grams. For a small group of 13 birds, find the margin of error for a 70% confidence interval for the average weights of these birds.

Group of answer choices

0.11 grams

0.06 grams

0.03 grams

0.52 grams

0.01 grams

Flag this Question

Question 61 pts

In a random sample of 41 professional actors, it was found that 15 were extroverts. Find a 99% confidence interval for p.

Group of answer choices

0.239 to 1.175

0.163 to 1.175

0.163 to 0.569

0.163 to 0.492

0.239 to 0.492

Flag this Question

Question 71 pts

Suppose the method of tree ring dating gave the following dates A.D. for an archaeological excavation site.

1,247 1,254 1,224 1,207 1,151 1,280 1,271 1,283 1,233

Find a0 95% confidence interval for the mean of all tree ring dates from this archaeological site.

Group of answer choices

1,206.7 to 1,271.1

1,207.4 to 1,271.1

1,208.6 to 1,270.4

1,208.6 to 1,269.2

1,207.4 to 1,270.4

Flag this Question

Question 81 pts

In a random sample of 50 professional actors, it was found that 36 were extroverts. Let p represent the proportion of all actors who are extroverts. Find a point estimate for p.

Group of answer choices

0.360

0.720

14.000

0.640

0.280

Solutions

Expert Solution

Solution :


Related Solutions

A random sample of 338 medical doctors showed that 168 had a solo practice. (a) Let...
A random sample of 338 medical doctors showed that 168 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. 99% of the confidence intervals created using this method would include the true proportion of...
A random sample of 330 medical doctors showed that 164 had a solo practice. (a) Let...
A random sample of 330 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 all confidence intervals would include the true proportion of physicians with solo...
A random sample of 328 medical doctors showed that 160 had a solo practice. (a) Let...
A random sample of 328 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 90% confidence interval for p. (Use 3 decimal places.) lower limit upper limit Give a brief explanation of the meaning of the interval. 10% of the confidence intervals created using this method would include the true proportion of...
A random sample of 340 medical doctors showed that 164 had a solo practice. (a) Let...
A random sample of 340 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. A) 90% of the all confidence intervals would include the true proportion of physicians with...
A random sample of 334 medical doctors showed that 174 had a solo practice. (a) Let...
A random sample of 334 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 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 true proportion of...
A random sample of 332 medical doctors showed that 176 had a solo practice. (a) Let...
A random sample of 332 medical doctors showed that 176 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 What is the margin of error based on a 98% confidence interval? (Use 3 decimal places.)
A random sample of 334 medical doctors showed that 174 had a solo practice. (a) Let...
A random sample of 334 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 95% confidence interval for p. (Use 3 decimal places.) lower limit upper limit Give a brief explanation of the meaning of the interval. 95% of the all confidence intervals would include the true proportion of physicians with...
A random sample of 324 medical doctors showed that 178 had a solo practice. (b) Find...
A random sample of 324 medical doctors showed that 178 had a solo practice. (b) Find a 98% confidence interval for p. (Use 3 decimal places.) lower limit upper limit What is the margin of error based on a 98% confidence interval? (Use 3 decimal places.)
Which medical practice model is referred to as a "solo practice"?
Which medical practice model is referred to as a "solo practice"?
(3 pts) A random sample of 100 freshman showed 10 had satisfed the university mathematics requirement...
(3 pts) A random sample of 100 freshman showed 10 had satisfed the university mathematics requirement and a second random sample of 50 sophomores showed that 12 had satisfied the university mathematics requirement. Enter answers below rounded to three decimal places. (a) The relative risk of having satisfied the university mathematics requirement for sophomores as compared to freshmen is (b) The increased risk of having satisfied the university mathematics requirement for sophomores as compared to freshmen is
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT