Question

A randomly selected sample of n =51 men in Brazil had an average lifespan of 59 years. The standard deviation was 10 years. Calculate a 98% confidence interval for the average lifespan for all men in Brazil.

Answer 1

Answer:

$59-2.40\frac{10}{\sqrt{51}}=55.639$    

$59+ 2.40\frac{10}{\sqrt{51}}=62.361$    

The 98% confidence interval would be given by (55.639;62.361)    

Step-by-step explanation:

Information given

$\bar X= 59$ represent the sample mean

$s= 10$ represent the sample deviation

$n= 51$ represent the sample size

Solution to the problem

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:

$df=n-1=51-1=50$

The Confidence is 0.98 or 98%, the significance would be $\alpha=0.02$ and $\alpha/2 =0.01$,and the critical value would be $t_{\alpha/2}=2.40$

And replacing we got:

$59-2.40\frac{10}{\sqrt{51}}=55.639$    

$59+ 2.40\frac{10}{\sqrt{51}}=62.361$    

The 98% confidence interval would be given by (55.639;62.361)