Question

A simple random sample from a population with a normal distribution of 103 body temperatures has x overbarequals98.90degrees Upper F and sequals0.62degrees Upper F. Construct a 95​% confidence interval estimate of the standard deviation of body temperature of all healthy humans.

Answer 1

Its 0.57 and 0.75

Explanation:

Answer 2

`98.90-1.984(0.62)/(√(103))=96.98`

`98.90+ 1.984(0.62)/(√(103))=100.82`

The 95% confidence interval would be given by (96.98;100.82)

Explanation:

Information given

`\bar X=98.90` represent the sample mean

`\mu` population mean (variable of interest)

s=0.62 represent the sample standard deviation

n=103 represent the sample size

Confidence interval

The confidence interval for the mean is given by the following formula:

`\bar X \pm t_(\alpha/2)(s)/(√(n))` (1)

The degrees of freedom are given by:

`df=n-1=103-1=102`

Since the Confidence is 0.95 or 95%, the significance is
`\alpha=0.05` and
`\alpha/2 =0.025`, and the critical value for this case would be
`t_(\alpha/2)=1.984`

And replacing we got:

`98.90-1.984(0.62)/(√(103))=96.98`

`98.90+ 1.984(0.62)/(√(103))=100.82`

The 95% confidence interval would be given by (96.98;100.82)