Question

In: Statistics and Probability

A sociologist sampled 200 people who work in computer related jobs, and found that 42 of...

A sociologist sampled 200 people who work in computer related jobs, and found that 42 of them have changed jobs in the past 6 months. Construct a 85% confidence interval for the proportion of those who work in computer-related jobs who have changed jobs in the past six months

Solutions

Expert Solution

Solution :

Given that,

n = 200

x = 42

Point estimate = sample proportion = = x / n = 42/200=0.21

1 -   = 1-0.21 =0.79

At 85% confidence level the z is ,

= 1 - 85% = 1 - 0.85 = 0.15

/ 2 = 0.075

Z/2 = Z0.075 = 1.44 ( Using z table )

Margin of error = E = Z/2   * ((( * (1 - )) / n)

= 1.44(((0.21*0.79) /200 )

E = 0.0415

A 85% confidence interval for population proportion p is ,

- E < p < + E

0.21-0.0415< p <0.21+ 0.0415

0.1685< p < 0.2515

The 85% confidence interval for the population proportion p is :0.1685, 0.2515

orchestra answered 3 years ago
Next > < Previous

Related Solutions

A sociologist is interested in surveying workers in computer-related jobs to estimate the proportion of such...
A sociologist is interested in surveying workers in computer-related jobs to estimate the proportion of such workers who have changed jobs within the past year. A)In the absence of preliminary data, how large a sample must be taken to ensure that a 95% confidence interval will specify the proportion to within ±0.02? Round up the answer to the nearest integer. Answer is not 2401 B) In a sample of 105 workers, 20 of them had changed jobs within the past...
A sociologist is interested in surveying workers in computer-related jobs to estimate the proportion of such...
A sociologist is interested in surveying workers in computer-related jobs to estimate the proportion of such workers who have changed the jobs within the past year. (a) In the absence of preliminary data, how large a sample must be taken to ensure that a 99% confidence interval will specify the proportion to within ±0.05? (b) Is the sample size obtained in (a) large enough to ensure that a 95% confidence interval will specify the proportion to within ±0.05? Explain. (c)...
What is the average family income of people who work in low-wage jobs? Without the resources...
What is the average family income of people who work in low-wage jobs? Without the resources of the EEOC, how likely is it a. that someone like Jamey could have per- sonally brought this case against Walmart, b. thatthecasewouldhavegonetotrial,and c. that Jamey could have engaged in an eleven-year litigation? here is the case study Pregnancy Discrimination at Walmart-Case Settled when the Baby Is 10 Years Old! Although the PDA has existed for many years and employers should therefore be well...
As a sociologist, you are interested in investigating the stress levels of people who live in...
As a sociologist, you are interested in investigating the stress levels of people who live in urban versus rural areas. You randomly select 80 urban residents and 70 rural residents to interview. Respondents from urban areas experience an average of 13.2 stressful experiences per week with a standard deviation of 3.2 times. Respondents from rural areas experience an average of 11.5 stressful experiences per week with a standard deviation of 2.1 times. We set the alpha level at 0.01. Please...
Suppose that people without jobs who were looking for work stop doing so. Other things the...
Suppose that people without jobs who were looking for work stop doing so. Other things the same, this makes the labor force will remain the same and the unemployment rate falls both the labor force and the unemployment rate remains the same both the labor force and the unemployment rate fall the labor force will fall while the unemployment rate remains the same
A national caterer determined that 36% of the people who sampled their food said that it...
A national caterer determined that 36% of the people who sampled their food said that it was delicious. A random sample of n=144 people is obtained from a large population. The 144 people are asked to sample the caterer's food. What is the probability that more than 34% thought the food was delicious? Question 20 options: A) 0.6915 B) 0.5987 C) 0.3085 D) 0.4013 Question 19 (1 point) A national caterer determined that 36% of the people who sampled their...
Out of 200 people sampled, 12 preferred Candidate A. Based on this, you will estimate what...
Out of 200 people sampled, 12 preferred Candidate A. Based on this, you will estimate what proportion of the voting population (pp) prefers Candidate A. Use technology to find a 90% confidence level, and give your answers as decimals, to four places. [_____ ,_______]
Out of 200 people sampled, 120 preferred Candidate A. Based on this, estimate what proportion of...
Out of 200 people sampled, 120 preferred Candidate A. Based on this, estimate what proportion of the voting population ( π π ) prefers Candidate A. Use a 95% confidence level, and give your answers as decimals, to three places.
Out of 200 people sampled, 156 preferred Candidate A. Based on this, estimate what proportion of...
Out of 200 people sampled, 156 preferred Candidate A. Based on this, estimate what proportion of the voting population ( p p ) prefers Candidate A. Use a 99% confidence level, and give your answers as decimals, to three places.
1) Out of 200 people sampled, 14 had kids. Based on this, construct a 95% confidence...
1) Out of 200 people sampled, 14 had kids. Based on this, construct a 95% confidence interval for the true population proportion of people with kids. Give your answers as decimals, to three places < p < 2) If n=450n=450 and ˆpp^ (p-hat) =0.91, find the margin of error at a 90% confidence level. Round to 4 places. z-scores may be rounded to 3 places or exact using technology. 3) In a recent poll, 110 people were asked if they...
ADVERTISEMENT
Subjects
ADVERTISEMENT
Latest Questions
ADVERTISEMENT