A simple random sample from a population with a normal distribution of 100 body temperatures has x 99.10°F and s = 0.61°F. Construct an 80 % confidence interval estimate of the standard deviation of body temperature of all healthy humans. Click the icon to view the table of Chi-Square critical values.

Answer)

Given confidence level is 80%

Variance = 0.61*0.61 = 0.3721

N = 100

Formula for confidence interval is

√{(n-1)*s^2}/√x^2(alpha/2) < sigma < √{(n-1)*s^2}/√x^2(1 - (alpha/2)}

Alpha = 1 - confidence level

S^2 sample.variance

N-1 = degrees of freedom

Degrees of freedom = n-1 = 99

Alpha = 1 - 0.8 = 0.2

For low end alpha/2 = 0.1

From chi square table, for 0.1 alpha and 99 df

X^2 = 117.4069

So low end.

= √{99*0.3721}/√117.4069}

= 0.5601

Upper end

Alpha = 1 - (alpha/2) = 0.9

For 0.9 alpha and 99 df

X^2 = 81.4493

Upper end = √{99*0.3721)/√81.4493 = 0.6725

Required interval is

0.5601 < sigma < 0.6725

= 0.56 < sigma < 0.67