Thread: Statistic confidence interval

hi, i am doing homework for my stat class and i am stuck on this one question. could you guys show me step by step how to solve this problem? the question is

Dr. Steven Wasserman, Dr. Philip Mackowiak, and Dr. Myron Levine of the University of Maryland conducted research on the body temperatures of healthy human adults. Fifty of the body temperatures taken from randomly selected subjects are provided below. Form a 98% confidence interval for the true average body temperature for all healthy adults, interpret the results. Does the interval contradict the belief that 98.6 is the average healthy body temperature, explain? Construct a hypothesis test also to determine if the status quo belief that the average healthy human body temp is 98.6 can be rejected using this data. Are the results from both methods consistent?

98.6 98.6 98.0 97.3 97.2
98.6 98.8 98.3 97.6 98.4
98.0 98.6 98.5 98.2 98.6
98.0 97.0 97.3 99.6 98.2
99.0 97.0 98.7 98.7 98.0
98.4 98.8 97.4 99.4 97.8
98.4 97.6 98.9 98.2 98.0
98.4 97.7 98.6 98.0 98.4
98.4 98.8 99.5 98.6 98.6
98.6 98.0 97.5 98.6 98.6

thank you

Originally Posted by cloudspear

hi, i am doing homework for my stat class and i am stuck on this one question. could you guys show me step by step how to solve this problem? the question is

Dr. Steven Wasserman, Dr. Philip Mackowiak, and Dr. Myron Levine of the University of Maryland conducted research on the body temperatures of healthy human adults. Fifty of the body temperatures taken from randomly selected subjects are provided below. Form a 98% confidence interval for the true average body temperature for all healthy adults, interpret the results. Does the interval contradict the belief that 98.6 is the average healthy body temperature, explain? Construct a hypothesis test also to determine if the status quo belief that the average healthy human body temp is 98.6 can be rejected using this data. Are the results from both methods consistent?

98.6 98.6 98.0 97.3 97.2
98.6 98.8 98.3 97.6 98.4
98.0 98.6 98.5 98.2 98.6
98.0 97.0 97.3 99.6 98.2
99.0 97.0 98.7 98.7 98.0
98.4 98.8 97.4 99.4 97.8
98.4 97.6 98.9 98.2 98.0
98.4 97.7 98.6 98.0 98.4
98.4 98.8 99.5 98.6 98.6
98.6 98.0 97.5 98.6 98.6

thank you

Where are you stuck? What have you been able to do? I for one do not plan to spend time getting the mean of this data ...... Have you?

Originally Posted by mr fantastic

Where are you stuck? What have you been able to do? I for one do not plan to spend time getting the mean of this data ...... Have you?

i got up to the getting the mean which is 98.28 i'm stuck on getting the standard divation. it is not given in the problem. could u tell me how to get the standard divation? thanks

Originally Posted by cloudspear

i got up to the getting the mean which is 98.28 i'm stuck on getting the standard divation. it is not given in the problem. could u tell me how to get the standard divation? thanks

I assume (hope) you used some form of technology to calculate the sample mean .....!! Can't that same technology give you the sample variance and hence standard deviation?

Originally Posted by mr fantastic

I assume (hope) you used some form of technology to calculate the sample mean .....!! Can't that same technology give you the sample variance and hence standard deviation?

i just add up all the number and i divide it by 50 and got 98.28 for the mean. so far i got x = 98.28, n = 50, CL = .98, alpha = .02 i'm doing it by hand and not using any technolgy because i will be test on this subject so i will need to be able to do it manually.

Originally Posted by cloudspear

i just add up all the number and i divide it by 50 and got 98.28 for the mean. so far i got x = 98.28, n = 50, CL = .98, alpha = .02 i'm doing it by hand and not using any technolgy because i will be test on this subject so i will need to be able to do it manually.

I can tell you now that no sane instructor (and probably not too many INsane instructors) will require you to calculate by hand the variance of a sample this size.