PLEASE EXPLAIN AND TEACH/ FAN+MEDAL

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.

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.

Question

PLEASE EXPLAIN AND TEACH/ FAN+MEDAL

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.

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.

anonymous · Answer

Using complete sentences, explain what these confidence intervals mean in the context of the problem.

dan815 · Answer

complete sentences eh

anonymous · Answer

more of a brief statement would be fine xD but i want to learn aswell

dan815 · Answer

A confidence interval is the interval that a certain event occurs, you need a certain precision associated with it, like 99% confidence interval means you are sure that 99% of the time your event occurs in this interval.

anonymous · Answer

okay.

anonymous · Answer

oh wait thats what its asking for.....

dan815 · Answer

yes

dan815 · Answer

you need to find the standard deviation that will cover 99% of the probable body temperatures

anonymous · Answer

help o.o i dunno Learning part xD

dan815 · Answer

okay well i have to ask you this first.. what grade are you in

anonymous · Answer

11

dan815 · Answer

because you can either use numerical integration or tables

anonymous · Answer

easiest way please

dan815 · Answer

okay then I think they expect you to just use the tables for standard deviation

anonymous · Answer

thats poop .

dan815 · Answer

your text book should have it on the last pages

anonymous · Answer

i actually do online school.

dan815 · Answer

a Z table or a standard deviation table and the percentages invovled

anonymous · Answer

yes i have a z chart

dan815 · Answer

okay so find what Z value you need for 99%

anonymous · Answer

so id go to 1 right?

anonymous · Answer

.16109

dan815 · Answer

no its gonna be like atleast more than 2 standard deviation

dan815 · Answer

i dont have a z score table

dan815 · Answer

you have to get your Z value and calcuate the temp
Z= (Temp - Mean Temp )/Standard Deviation

dan815 · Answer

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