ch.8 confidence interval mean

Question

question 9

The body temperatures in degrees Fahrenheit of a sample of adults in one small town are:

99.3

98.5

97.8

99.9

98.9

98.1

99.4

96.8

97.1

97.4

97.7

96.3

Assume body temperatures of adults are normally distributed. Based on this data, find the 98% confidence interval of the mean body temperature of adults in the town. Enter your answer as an open-interval (i.e., parentheses) accurate to 3 decimal places. Assume the data is from a normally distributed population.

98% C.I. =   

Answer 1

Cudos for the person asking the question. Well formulated with all the needed information.

I'm going to solve this using R programing language

sample =c( 99.3,98.5,97.8,99.9,98.9,98.1,99.4,96.8,97.1,97.4,97.7, 96.3)

n=length(sample)

sd = sd(sample)

xbar = mean(sample)

UCB = qnorm(.01,mean=xbar,sd=sd/sqrt, lower.tail=FALSE)

LCB = qnorm(.01,mean=xbar,sd=sd/sqrt, lower.tail=TRUE)

we get n =. 12, sample sd = 1.12169 xbar = 98.1 alpha/2 = .01

Confidence Interval =. (97.347 98.853)



Small correction. In last sentence, the verb is 'are', data are plural , so "Assume the data are from a normally distributed population. "