Asked by Ryler
Help me solve this please. One of the few on my list that I had no idea how to solve.
A medical researcher is interested in the prenatal care received by pregnant women in inner cities. She interviews 35 randomly selected women with children on the streets of Baltimore and finds that the average number of gynecological checkups per pregnancy was 3, with a standard deviation of 1. Using a 95% level of confidence, estimate the population mean number of gynecological visits per pregnancy.
In the end I got the confidence level to be around 1.96 - 4.04. I also got 1.96 to 4.03 at one point. But when I check the answer on the back of the book it says 2.65-3.35. Can someone go step by step on how to solve this.
A medical researcher is interested in the prenatal care received by pregnant women in inner cities. She interviews 35 randomly selected women with children on the streets of Baltimore and finds that the average number of gynecological checkups per pregnancy was 3, with a standard deviation of 1. Using a 95% level of confidence, estimate the population mean number of gynecological visits per pregnancy.
In the end I got the confidence level to be around 1.96 - 4.04. I also got 1.96 to 4.03 at one point. But when I check the answer on the back of the book it says 2.65-3.35. Can someone go step by step on how to solve this.
Answers
Answered by
PsyDAG
95% = mean ± 1.96 SEm
SEm = SD/√n = 1/√35 = 1/5.92 = .17
.17 * 1.96 = .33
95% = mean ± 1.96 SEm = 3 ± .33 = 2.67-3.33
Rounding error?
SEm = SD/√n = 1/√35 = 1/5.92 = .17
.17 * 1.96 = .33
95% = mean ± 1.96 SEm = 3 ± .33 = 2.67-3.33
Rounding error?
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