calculate: PART A PART B PART C
PART A:
Out of 300 people sampled, 204 preferred Candidate A.
Based on this, find a 90% confidence level for the true proportion of the voting population (pp) prefers Candidate A.
Give your answers as decimals, to three places.
< pp <
PART B: In a sample of 390 adults, 289 had children. Construct a
95% confidence interval for the true population proportion of
adults with children.
Give your answers as decimals, to three places
< p <
PART C:
Out of 400 people sampled, 140 preferred Candidate A. Based on this, estimate what proportion of the voting population (p) prefers Candidate A.
Use a 95% confidence level, and give your answers as decimals, to
three places.
< p <
calculate: PART A PART B PART C PART A: Out of 300 people sampled, 204 preferred...
Out of 500 people sampled, 235 preferred Candidate A. Based on this, find a 95% confidence level for the true proportion of the voting population (P) prefers Candidate A. Give your answers as decimals, to three places <pく 20 Points possible:1
Out of 100 people sampled, 44 preferred Candidate A. Based on this, estimate what proportion of the voting population (pp) prefers Candidate A. Use a 95% confidence level, and give your answers as decimals, to three places. __________< p < __________
NOTE VOTE Out of 300 people sampled, 249 preferred Candidate A. Based on this, estimate what proportion of the voting population (TT) prefers Candidate A. Use a 95% confidence level, and give your answers as decimals, to three places. <t< If n=560 and p (p-hat) =0.23, find the margin of error at a 90% confidence level Give your answer to three decimals License Points possible: 1 This is attempt 1 of 1.
QUESTION PART A: Out of 300 people sampled, 219 had kids. Based on this, construct a 99% confidence interval for the true population proportion of people with kids. Give your answers as decimals, to three places ____< p < ____ QUESTION PART B: Express the confidence interval 154.5±42.9154.5±42.9 in interval form.
In a recent poll, 350 people were asked if they liked skiing, and 55% said they did. Find the margin of error of this poll, at the 90% confidence level. As in the reading, in your calculations: --Use z1.645 for a 90% confidence interval -Use z=2 for a 95% confidence interval -Use z-2.576 for a 99% confidence interval Give your answer rounded to three decimal places. If n=560 and p (p-hat) = 0.7, construct a 99% confidence interval. As in...
all parts, show work please a) We wish to estimate what percent of adult residents in a certain county are parents. Out of 200 adult residents sampled, 152 had kids. Based on this, construct a 95% confidence interval for the proportion ?? of adult residents who are parents in this county. Give your answers as decimals, to three places. _____ < ? < _____ b) Out of 200 people sampled, 150 preferred Candidate A. Based on this, estimate what proportion...
Out of 200 people sampled, 68 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
Out of 400 people sampled, 264 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
Out of 100 people sampled, 51 had kids. Based on this, construct a 90% confidence interval for the true population proportion of people with kids. Give your answers as decimals, to three places ? < p < ?
Show IntrodInstructions Out of 100 people sampled, 35 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 four places <p Get Help Points possible: 1 This is attempt 1 of 2 Sub