Question

A simple random sample from a population with a normal distribution of

9797

body temperatures has

x overbarxequals=98.7098.70degrees Upper F°F

and

sequals=0.670.67degrees Upper F°F.

Construct

a

99​%

confidence interval estimate of the standard deviation of body temperature of all healthy humans.

Answer 1

ANSWER:

Given that,

A simple random sample from a population with a normal distribution of 97 body temperatures has x overbar=98.70 degrees Upper F° and s = 0.67 degrees Upper F°.Construct a 99% confidence interval estimate of the standard deviation of body temperature of all healthy humans.

Sample size = n = 97

= 98.70

s = 0.67

C = 99% = 99/100 = 0.99

= 1-C = 1-0.99 = 0.01

/2 = 0.01/2 = 0.005

1-(/2) = 1-(0.01/2) = 0.995

Degree of freedom = df = n-1 = 97-1 = 96

Critical values,

= = 64.063327 and

= = 135.433049

Confidence interval

(n-1) / < < (n-1) /

(97-1)0.67^2 / 135.433049 < < (97-1)0.67^2 / 64.063327​

0.318197074 < < 0.672684389

sqrt(0.318197074 ) < < sqrt(0.672684389)

0.564089 < < 0.820173

0.5641  < < 0.8202  (Rounded to four decimal place)

The confidence interval estimate is 0.5641  < < 0.8202

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