Question

In: Statistics and Probability

1. Assume that a sample is used to estimate a population proportion p. Find the margin...

1. Assume that a sample is used to estimate a population proportion p. Find the margin of error M.E. that corresponds to a sample of size 224 with 25 successes at a confidence level of 99.8%.
M.E. = ___%

2. Assume that a sample is used to estimate a population proportion p. Find the margin of error M.E. that corresponds to a sample of size 310 with 46.1% successes at a confidence level of 99.5%.
M.E. = ____%

3.Assume that a sample is used to estimate a population proportion p. Find the 98% confidence interval for a sample of size 251 with 161 successes. Enter your answer as an open-interval (i.e., parentheses) using decimals (not percents) accurate to three decimal places.

98% C.I. = ____

Solutions

Expert Solution

(1) proportion p = (success)/(total sample size) = 25/224 = 0.1116

sample size n = 224

Significance level = 1-(99.8/100) = 1-0.998 = 0.002

using z distribution table for 0.002 significance level, we get z critical value = 3.09

Formula for margin of error is given as

setting the values, we get

converting to %, we get 0.065*100 = 6.5%

Therefore, required margin of error is 6.5% (rounded to 1 decimal) or 6.50%  (rounded to 2 decimal)

(2) proportion p = 46.1/100 = 0.461

sample size n = 310

Significance level = 1-(99.5/100) = 1-0.995 = 0.005

using z distribution table for 0.005 significance level, we get z critical value = 2.81

Formula for margin of error is given as

setting the values, we get

converting to %, we get 0.0796*100 = 7.96 or 8.0 %

Therefore, required margin of error is 8.0% (rounded to 1 decimal) or 7.96% (rounded to 2 decimal)

(3) it is given that there are 161 successes out of 251

So, proportion = 161/251

significance level = 1-(confidence level)/100 = 1-(98/100) = 0.02

Using z distribution table for 0.02 level of significance, we get z critical value = 2.33

sample size is n= 251

Using the formula for the confidence interval, we can write

setting the given values, we get

this gives us

or, we can write it as

So, we get Confidence interval = (0.571,0.712)

Therefore, required 98% confidence interval is (0.571,0.712) (rounded to 3 decimals)


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