Question

According to Thomson Financial, last year the majority of companies reporting profits had beaten estimates. A...

According to Thomson Financial, last year the majority of companies reporting profits had beaten estimates. A sample of 162 companies showed that 98 beat estimates, 29 matched estimates, and 35 fell short. (a) What is the point estimate of the proportion that fell short of estimates? If required, round your answer to four decimal places. pshort = .2160 (b) Determine the margin of error and provide a 95% confidence interval for the proportion that beat estimates. If required, round your answer to four decimal places. ME = .0753 (c) How large a sample is needed if the desired margin of error is 0.05? If required, round your answer to the next integer. I need your help solving C.

3 0
Add a comment Improve this question Transcribed image text
Answer #1

on 162 Best eshmaț ej -98 n: 162 marched estimate-29 ghor35 ell shor o etfmato O. 22 62Page thot beat imate 6-2 Alco Ko.ofe 46 rom jable) 2 62 ME 0 045Date Page f上 . → (0.6049 + 0.0753) ー) | ( 0 : 5296 , o .6802 ) Gven ME-0.0S 162 2 36

Add a comment
Know the answer?
Add Answer to:
According to Thomson Financial, last year the majority of companies reporting profits had beaten estimates. A...
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for? Ask your own homework help question. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions
  • 1. A. According to Fidelity Investments, through last month, the majority of companies reporting profits had...

    1. A. According to Fidelity Investments, through last month, the majority of companies reporting profits had beaten estimates. A sample of 160 companies showed that 96 beat estimates, 38 matched estimates, and 26 fell short of estimates. Using this information, how large a sample is needed to estimate the true proportion of all companies that had beaten estimates if you want to be 97% certain of being within 2% of the true proportion? Answer: n = B. Customers arrive at...

  • comfidence interval and margin of error SAGE MINDTAP WULL HOW Fly 10t? Search this course rk...

    comfidence interval and margin of error SAGE MINDTAP WULL HOW Fly 10t? Search this course rk 5: Confidence intervals anment Homework 5: Confidence Intervals Asment Score: 89.18% Save Submit Assignment for Grading Question 4 of 12 Check My Work eBook on Financial, last year fell short According to Thomson Financial, last year the majority of companies reporting profits had beaten estimates. A sample of 162 companies showed that 102 beat estimates, 29 matched estimates, and 31 fell short. (a) What...

  • (COMPUTE AND INTERPRET CONFIDENCE INTERVAL ESTIMATES) . According to the U.S. Census Bureau, 43% of men...

    (COMPUTE AND INTERPRET CONFIDENCE INTERVAL ESTIMATES) . According to the U.S. Census Bureau, 43% of men who worked at home were college graduates. In a sample of 500 women who worked at home, 162 were college graduates. Construct a 98% confidence interval for the proportion of women who work at home who are college graduates. Find the critical values, find E , the margin of error, then compute a nd interpret your interval estimate with a full sentence. Critical value:...

  • Question 5 Score on last try: 3 of 8 pts. See Details for more. Try a...

    Question 5 Score on last try: 3 of 8 pts. See Details for more. Try a similar question You can retry this question below Population proportion: unknown Sample 1 Sample 1 proportion: P = 0.78 Sample 1 size: n = 144 1. Using Sample 1, give a point estimate of the population proportion. .81 X 2. Using Sample 1, give an interval estimate of the population proportion with 98% confidence. "With 98% confidence, the population proportion is between 106 X...

  • Constructing Confidence Intervals, Part 1: Estimating Proportion Assume that a sample is used to estimate a...

    Constructing Confidence Intervals, Part 1: Estimating Proportion Assume that a sample is used to estimate a population proportion p. Find the margin of error E that corresponds to the given statistics and confidence level: In a random sample of 200 college students, 110 had part-time jobs. Find the margin of error for the 98% confidence interval used to estimate, for the entire population of college students, the percentage who have part-time jobs. Round your answer to three decimal places. Please...

  • For many years businesses have struggled with the rising cost of health care. But recently, the...

    For many years businesses have struggled with the rising cost of health care. But recently, the increases have slowed due to less inflation in health care prices and employees paying for a larger portion of health care benefits. A recent Mercer survey showed that 52% of U.S. employers were likely to require higher employee contributions for health care coverage. Suppose the survey was based on a sample of 400 companies. Compute the margin of error and a 95% confidence interval...

  • Please help eBook Fewer young people are driving. In 1983, 87% of 19-year-olds had a driver's...

    Please help eBook Fewer young people are driving. In 1983, 87% of 19-year-olds had a driver's license. Twenty-five years later that percentage had dropped to 75% (University of Michigan Transportation Research Institute website, April 7, 2012). Suppose these results are based on a random sample of 1200 19-year-olds in 1983 and again in 2008. a. At 95% confidence, what is the margin of error and the interval estimate of the number of 19-year-old drivers in 1983? Round your intermediate answers...

  • A random sample of 320 medical doctors showed that 180 had a solo practice. (a) Let...

    A random sample of 320 medical doctors showed that 180 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 all confidence intervals would include the true proportion of physicians with solo...

  • Fewer young people are driving. In 1983, 87% of 19-year-olds had a driver's license. Twenty-five years...

    Fewer young people are driving. In 1983, 87% of 19-year-olds had a driver's license. Twenty-five years later that percentage had dropped to 75% (University of Michigan Transportation Research Institute website, April 7, 2012). Suppose these results are based on a random sample of 1200 19-year-olds in 1983 and again in 2008 a. At 95% confidence, what is the margin of error and the interval estimate of the number of 19-year-old drivers in 1983? Round your intermediate answers to four decimal...

  • confidence level and sample data to find confidence interval for estimating a population round your answer...

    confidence level and sample data to find confidence interval for estimating a population round your answer to the same number of decimal places as a sample Question 2 2 pts Assume that a sample is used to estimate a population proportion p. Find the margin of error E that corresponds to the given statistics and confidence level. Round the margin of error to four decimal places. 90% confidence; n 390, x-146

ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT