Question

A random sample of 10 independent healthy people showed the following body temperatures (in degrees Fahrenheit) and summary statistics:

Temps

98.50

98.20

99.00

96.30

98.30

98.70

97.20

99.10

98.70

97.20

Average

98.12

Std Dev

0.92

Compute a 95% CI for the average body temperature of healthy people.

Answer 1

showed. temperatures of of ith selected 10 independent healthy booble people. In this preoblem body xi= body temperature average () = 98.12. Std. Der (6) = 0.92. no. of samble () = 10 distribution therefore, Here but 95% we assume that data coming from Normal since o (Population std. devi) is not known confidence interval is given by - x- 2 tri, a ž + 1 hora interval. Therefore since our objective is to find 95% confidence = 0.05 = 0.025 movegin error Hence, confidence interval is given by. e to cook . r +9,0.025 = 2.2621 I l from t-table. I J50 0.658 confidence interval : [ 7 - 0-658, Z+0.6587 - [ 97.462 98.418] cans) concept: Normal population. standard deviation. X., X2...xn be reandom sample from with unknown mean & unknown le *,*2 - - n N(M,O2) H=unknown unknown Then, X-M NN (0,7). 5/5 → X-t ~ tn-lain t-distribution with def = (n-1)