Carlisle conducted an experiment to determine if the there is a difference in mean body temperature for men and women. He found that the mean body temperature for men in the sample was 97.9 with a population standard deviation of 0.57 and the mean body temperature for women in the sample was 98.6 with a population standard deviation of 0.55.

Question

anonymous · Answer

Assuming the population of body temperatures for men and women is normally distributed, calculate the 99% 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.

anonymous · Answer

@Opcode

anonymous · Answer

The question asks for the confidence interval for the men not for the individual values, thus 
standard error of the mean =  s.d./sqrt(n) 
is the correct measure of variability and we assume means are distributed normally..

anonymous · Answer

For men, the pdf is
$$
f(x)=0.699899 e^{-1.53894
   (x-97.9)^2}
$$
For a=1.47, we have
$$
\int_{97.9\, -a}^{a+97.9}
   0.699899 e^{-1.53894
   (x-97.9)^2} \, dx=0.99009
$$
Which means that 99% of men have their body temperature between 96.43 and 99.37

anonymous · Answer

I hope that @douglaswinslowcooper will help you with the margin of error.

anonymous · Answer

how would you find out the women?

anonymous · Answer

@douglaswinslowcooper can you please show how you'd find this for women?