In: Statistics and Probability
The following data show the body temperatures (in degrees Fahrenheit) from a random sample of 12 independent healthy people in the United States.
98.5 98.2 99.1 96.6 98.1 98.7 97.5 99.0 97.4 98.3 97.8 98.2
What would be the estimate of the average body temperature of healthy people in the United States.
Find a 95% confidence interval for the average body temperature of healthy people in the United States.
Explain carefully the interval you found in part (b) means.
can you please show proof while figuring out the problems
x | (x-xbar)^2 | |
98.5 | 0.146944 | |
98.2 | 0.966944 | |
99.1 | 2.300278 | |
96.6 | 0.000278 | |
98.1 | 0.340278 | |
98.7 | 0.380278 | |
97.5 | 0.780278 | |
99 | 0.513611 | |
97.4 | 0.033611 | |
98.3 | 0.100278 | |
97.8 | 0.006944 | |
98.2 | 1164853 | |
sum | 1177.4 | 1164858 |
Mean(x)=xbar=sum(x)/n | 98.11667 |
standard deviation(s)=sum(x-xbar)^2/n-1 | 0.712018 |
n | 12 |
for 95 % confidence level with degree of freedom (n-1)=11 | |
=1-c%=1-0.05 | 0.05 |
degrres of freedom | 11 |
t=critical value obtain using t-table with corresponding df=n-1= | 2.200985 |
Margin of error =t*s/sqrt(n) | 0.452395 |
LCL=xbar-ME | 97.66427 |
UCL=xbar+ME | 98.56906 |
estimate of the average body temperature of healthy people in the United States.is xbar=98.12
#95% confidence interval for the average body temperature of healthy people in the United States. is
(xbar-ME,xbar+ME)
(97.67,98.57)----95 % CI
#One can be 95% confident that the mean body temperature of healthy people in US is between 97.67 and 98.57