3) assume that a sample is used to estimate a population proportion p. Find the

Question

3) 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. 3) 90% confidence; the sample size is 1510 , of which 35 % are successes

4) use the given degree of confidence and sample data to construct a confidenceinterval for the population proportion p. n=195, x=162; 95% confidence

5) Thirty randomly selected students took the calculus final. If the sample mean was 95 and the standard deviation was 6.6, construct a 99% confidence interval for the mean score of all students.The principal randomly selected six students to take an aptitude test. Their

scores were: 88.0 84.174.9 83.2 83.7 85.5

Determine a 90% confidence interval for the mean score for all students.

Explanation / Answer

margin of error = z *sqrt(pq/n)

90 % confidence

z = 1.645

n = 1510

hence

margin of error E = 1.645*sqrt(0.35*0.65/1510) = 0.02019147

4) for 95 % CI

z = 1.96

p^ =X/n = 162/195 = 0.83076,

q^ = 1 -0.83076 = 0.169230

margin of error E = 1.96 *sqrt(0.169230*0.83076/195) = 0.0526278

hence CI

=(0.83076 -0.0526278 , 0.83076+ 0.0526278)

=(0.7781322,0.8833878)

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