Construct a confidence interval for the population proportion p.
Sample size, n=256, success number, x=130, 90% confidence.
Solution :
Given that,
Point estimate = sample proportion = = x / n = 130 / 256 = 0.508
1 - = 1 - 0.508 = 0.492
Z/2 = 1.645
Margin of error = E = Z / 2 * (( * (1 - )) / n)
= 1.645 * (((0.508 * 0.492) / 256)
Margin of error = E = 0.051
A 90% confidence interval for population proportion p is ,
- E < p < + E
0.508 - 0.051 < p < 0.508 + 0.051
0.457 < p < 0.559
The 90% confidence interval for the population proportion p is : 0.457 , 0.559
Construct a confidence interval for the population proportion p. Sample size, n=256, success number, x=130, 90%...
Construct a confidence interval for the population proportion p. Sample size, n=256, success number, x=130, 90% confidence.
Question 1 1. Construct a confidence interval for the population proportion p. 1) Sample size, n=256, success number, x = 130; 95% confidence O 0.425 <p < 0.647 O 0.405 <p <0.667 O 0.404 < p < 0.668 O 0.447 <p <0.569 Question 2
Determine the sample size n needed to construct a 90% confidence interval to estimate the population proportion when p = 0.65 and the margin of error equals 7%. n= (Round up to the nearest integer.)
Construct a 90% confidence interval to estimate the population proportion with a sample proportion equal to 0.44 and a sample size equal to 100. A 90% confidence interval estimates that the population proportion is between a lower limit of blank and an upper limit of. (Round to three decimal places as needed.)
(22) The 99% confidence interval for the TRUE PROPORTION of success for a population is (0.318, 0.462). The random sample size is 300. (i) Please determine the SAMPLE proportion of success. (ii) Please determine the MARGIN FOR ERROR. (ii) Please determine the NUMBER OF SUCCESSFUL OUTCOMES. (23) The 90% confidence interval for the ACTUAL MEAN of a given population is (84, 90 ), via a "z" analysis. The random sample size is 81. (i) Please determine the (A) SAMPLE AVERAGE...
If n = 310 and X = 248, construct a 90% confidence interval for the population proportion, p. Give your answers to three decimals
Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p. Round to three decimal palces Of 98 adults selected random from one town, 68 have health insurance Find a 90% confidence interval adults in he own who have heal e proportion on r e a ns rance 0585 < p < 0 802 B. 0.617<p<0.770 A. C. 0603 p<0.785 D. 0.574p<0.814 Use the given degree of confidence and sample data to...
If n=300 and X=120, construct a 90% confidence interval estimate for the population proportion. Round to four decimal places as needed
. Construct a confidence interval of the population proportion at the given level of confidence. x =300, n = 1200, 90% confidence.
Construct a confident interval of the population proportion at the given level of confidence. X=120 n=300 90% Confidence Thanks! 8 of 9 (0 Construct a confidence interval of the population proportion at the given level of confidence xz 120, n: 300, 90% confidence The lower bound is The upper bound is (Round to three decimal places as needed) Enter your answer in each of the answer boxes. O Type here to search