Question

In: Statistics and Probability

In a random sample of 61 professional actors, it was found that 48 were extroverts. Find...

In a random sample of 61 professional actors, it was found that 48 were extroverts. Find a 99% confidence interval for p.

Expert Solution

Solution :

Given that,

n = 61

x = 48

Point estimate = sample proportion = = x / n = 48 / 61 = 0.787

1 - = 1 - 0.787 = 0.123

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

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

= 2.576 * ( $\sqrt$ ((0.787 * 0.213) / 61)

= 0.135

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.787 - 0.135 < p < 0.787 + 0.135

0.652 < p < 0.922

The 99% confidence interval for the population proportion p is : 0.652 , 0.922

orchestra answered 2 years ago

In a random sample of 61 professional actors, it was found that 22 were extroverts. Find...

In a random sample of 61 professional actors, it was found that 22 were extroverts. Find a 95% confidence interval for p, the population proportion of actors that are extroverts

1) In a random sample of 61 professional actors, 22 were extroverts. a. Comment on the...

1) In a random sample of 61 professional actors, 22 were extroverts. a. Comment on the requirements of a confidence interval. b. Find a 90% confidence interval for p. 2) If the average height x is calculated from a random sample of 135 adult Americans, find the probability that the mean height of the sample is between 68 and 70 inches. 3) A food distributor is attempting to deal with complaints that an unacceptable percentage of eggs transported by the...

In a random sample of 65 professional actors, it was found that 43 were extroverts. (a)...

In a random sample of 65 professional actors, it was found that 43 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

In a random sample of 63 professional actors, it was found that 45 were extroverts. (a)...

In a random sample of 63 professional actors, it was found that 45 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 Give a brief interpretation of the meaning of the confidence interval you have found. We are 5% confident that the true...

Professional Golfers’ Earnings Two random samples of earnings of professional golfers were selected. One sample was...

Professional Golfers’ Earnings Two random samples of earnings of professional golfers were selected. One sample was taken from the Professional Golfers Association, and the other was taken from the Ladies Professional Golfers Association. At a=0.10, is there a difference in the means? The data are in thousands of dollars. Use the critical value method with tables. PGA 1147, 1344, 9188, 5687, 10508, 4910, 8553, 7573, 375 LPGA. 48, 76, 863, 100, 1876, 2029, 4364 Assume the variables are normally distributed...

A random sample of 500 students were selected and it was found that 345 of them...

A random sample of 500 students were selected and it was found that 345 of them were satisfied with their field. Construct a 95% confidence interval for the true proportion of students who are satisfied with their field. (i)Identify n, p̂, q̂, and ??/2. ? = p̂ = q̂ = ??/2 = (ii)Calculate the margin of error, E. (iii)Construct the 95% confidence interval

In a random sample of 506 judges, it was found that 283 were introverts. P= .5593...

In a random sample of 506 judges, it was found that 283 were introverts. P= .5593 a. Find a 99% confidence interval for p. (Round your answers to two decimal places.) a1. lower limit a2. upper limit b. Do you think the conditions np > 5 and nq > 5 are satisfied in this problem? Explain why this would be an important consideration. A. Yes, the conditions are satisfied. This is important because it allows us to say that p̂...

A random sample of 150 WCC students found that 68 of them were Latino/a. a) At...

A random sample of 150 WCC students found that 68 of them were Latino/a. a) At the 5% level of significance can we prove that over 40% of all WCC students are Latino/a? State and test appropriate hypotheses. State conclusions. b) Find the p-value of the test in (a). c) If an error was made in (a), what type was it?

In a random sample of 65 teenage males, 42 were found to be overweight. a) Construct...

In a random sample of 65 teenage males, 42 were found to be overweight. a) Construct then state a 90% confidence interval from this sample result for the true proportion of teenage males that are overweight. b) Interpret your interval in this context. c) From your interval, are significantly less than 80% of teenage males? Explain briefly.

A random sample of 100 voters found that 46% were going to vote for a certain...

A random sample of 100 voters found that 46% were going to vote for a certain candidate. Find the 90% confidence interval for the population proportion of voters who will vote for that candidate. A. 38.7% < p < 53.3% B. 37.8% < p < 54.2% C. 41.9% < p < 50.1% D. 39.6% < p < 52.4%