9:50 "4G 28.0KB/SCompute the populationproportion interval estimategiven n, p, and the confidencelevel, *n =300, p = 0.40, 95% confidencep = (0.483 < P < 0.649)H = (0.556 < P <0.649)Ο μ- (0.571< μ< 0.649)Compute the populationproportion interval estimategiven n, p, and the confidencelevel, *n = 420, p = 0.61, 90% confidenceO H = (0.483 < P < 0.649)O p = (0.556 < P <0.649)O H = (0.571 < µ < 0.649)->

Name: 9:50 "4G 28.0KB/SCompute the populationproportion interval estimategi
Uploaded: 2024-03-20T07:07:00.000Z
Channel: Bartleby
Description: 9:50 "4G 28.0KB/SCompute the populationproportion interval estimategiven n, p, and the confidencelevel, *n =300, p = 0.40, 95% confidencep = (0.483 < P < 0.649)H = (0.556 < P <0.649)Ο μ- (0.571< μ< 0.649)Compute the populationproportion interval est

Note: Please repost 2nd question as i can answer only 1 question as per my policy. Given data,…

Answered: Compute the population proportion…

Math

Statistics

Compute the population proportion interval estimate given n, p , and the confidence level. * 300, p = 0.40, 95% confidence

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

A 30 month study is conducted to determine the difference in rates of accidents per month between two departments in an assembly plant. If a sample of 12 from the first department averaged 12.3 accidents per month with a standard deviation of 3.5, and a sample of 9 from the second department averaged 7.6 accidents per month with standard deviation of 3.4, find: The t-value for a 95% confidence interval is: A. 2.09302 B. 1.724718 C. 2.08596 D. 1.729133 E. none of the preceding
In random, independent samples of 225 adults and 200 teenagers who watched a certain television show, 112 adults and 138 teens indicated that they liked the show. Let p1 be the proportion of all adults watching the show who liked it, and let p2 be the proportion of all teens watching the show who liked it. Find a 95% confidence interval for −p1p2. Then find the lower limit and upper limit of the 95% confidence interval. Carry your intermediate computations to at least three decimal places. Round your responses to at least three decimal places. a.) The Lower Limit is: b.) The Upper Limit is:
From a large number of actuarial exam scores, a random sample of 275 scores is selected, and it is found that 206 of these 275 are passing scores. Based on this sample, find a 90% confidence interval for the proportion of all scores that are passing. Then complete the table below. Carry the intermediate computations to at least three decimal places. Round the answers to two decimal places. (a) What is the lower limit of the 90% confidence interval? (b) What is the upper limit of the 90% confidence interval?
A random sample of n measurements was selected from a population with unkown mean μ and standard deviation σ =20 for each of the situations in parts a through d. Calculate a 99% confidence interval for μ for each of these situations. a. n=50, x̅=25 b. n=250, x̅=103 c. n=100, x̅=13 d. n=100, x̅=4.82 e. Is the assumption that the underlying population of measurements is normally distributed necessary to ensure the validity of the confidence intervals in parts a through d? Explain.
A sample of n=17n=17 data values randomly selected from a normally distributed population has standard deviation s=6.8s=6.8. Construct a 95% confidence interval for the population standard deviation. Round your endpoints to one decimal place.
In a study of maternal cigarette smoking and bone density in newborns, 77 infants of mothers who smoked had a mean bone mineral content of 0.098 g/cm3 (s1 = 0.026 g/cm3). The 161 infants whose mothers did not smoke had a mean bone mineral content of 0.095 g/cm3 (s2 = 0.025 g/cm3). Calculate the 95% confidence interval for µ1 - µ2
A random sample of size n1=25, taken from a normal population with a standard deviation σ1=5, has a mean x̄1=80. A second random sample of size n2=36, taken from a different normal population with a standard deviation σ2=3, has a mean x̄2=75. Find a 99% confidence interval for μ1-μ2.
A simple random sample of 49 8th graders at a large suburban middle school indicated that 82% of them are involved with some type of after school activity. Find the 99% confidence interval that estimates the proportion of them that are involved in an after school activity. a) (0.679, 0.761) b) (0.679, 0.961) c) (0.579, 0.911) d) (0.599, 0.961) e) (0.729, 0.734) f) None of the above
11) A journal article reports that a sample of size 5 was used as a basis for calculating at 95% CI for the true average natural frequency (Hz) of delaminated beams of a certain type. The resulting interval was (229.764, 233.504). Assume the population is normal. You decide that a confidence level of 99% would be more appropriate than a 95% level used. What are the limits of the 99% interval?
A random sample of size 30 from a normal population yields x̅ = 69 and s = 4. The lower bound of a 95 percent confidence interval is (Round off upto 2 decimal places).
An apple farmer inspects a large random sample of his apples and finds that 20% contain excellent worms. He wants to charge his customers extra, and claims that the 90% confidence interval for the population pro- portion of apples that contain worms has a margin of error of at most 0.01. At least how many apples did the farmer inspect?
A soft-drink manufacturer purchases aluminum cans from an outside vendor. A random sample of 70 cans is selected from a large shipment, and each is tested for strength by applying an increasing load to the side of the can until it punctures. Of the 70 cans, 51 meet the specification for puncture resistance. a) Find the sample size needed for a 90% confidence interval to specify the proportion to within ±0.06. Round up the answer to the nearest integer. b) If a 90% confidence interval is computed each day for 300 days, what is the probability that more than 280 of the confidence intervals cover the true proportions? Round the answer to three decimal places.