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Mathematics 11 Online
OpenStudy (anonymous):

FAN+MEDAL+Thankyou letter Assuming the population of body temperatures for men and women is normally distributed, calculate the 98% confidence interval and the margin of error for the mean body temperature for both men and women. Using complete sentences, explain what these confidence intervals mean in the context of the problem.

OpenStudy (cj49):

iz diz question relatd to mean diff in d body temp??

OpenStudy (anonymous):

i guess

OpenStudy (anonymous):

@midhun.madhu1987 @ilikemath50

OpenStudy (anonymous):

swave conducted an experiment to determine if there is a difference in the mean body temperature between men and women. He found that the mean body temperature for a sample of 100 men was 91.1 with a population standard deviation of 0.52 and the mean body temperature for a sample of 100 women was 97.6 with a population standard deviation of 0.45. Assuming the population of body temperatures for men and women is normally distributed, calculate the 98% confidence interval and the margin of error for the mean body temperature for both men and women. Using complete sentences, explain what these confidence intervals mean in the context of the problem.

OpenStudy (anonymous):

@KlOwNlOvE

OpenStudy (anonymous):

Sorry I don't exactly understand the question, it looks poorly written to me. But I wish you good luck on this question. :)

OpenStudy (anonymous):

ughhh

OpenStudy (sedatefrog712):

so much reading

OpenStudy (anonymous):

9 lines is much to read?

OpenStudy (anonymous):

for him it is hes a lil special;)

OpenStudy (sedatefrog712):

I am

OpenStudy (sedatefrog712):

don't hate me because im beutiful

OpenStudy (cj49):

r u familiar with empirical rule??

OpenStudy (cj49):

Men's CI x-bar = 91.1 ME = 2.3263*0.52 = 1.210 98% CI:: 91.1-1.21 < u < 91.1+1.21 98% CI: 89.89 < u < 92.31 Women's CI x-bar = 97.6 ME = 2.3263*0.45 = 1.05 98% CI: 97.6-1.05 < u < 97.6+1.05 98% CI: 96.55 < u < 98.65

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