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

Answer:

$98.90-1.984\frac{0.62}{\sqrt{103}}=96.98$    

$98.90+ 1.984\frac{0.62}{\sqrt{103}}=100.82$    

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

Step-by-step 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:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$   (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:

$98.90-1.984\frac{0.62}{\sqrt{103}}=96.98$    

$98.90+ 1.984\frac{0.62}{\sqrt{103}}=100.82$    

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