Question

You are studying the ages of people afflicted with acne. You collected a random sample of 49 people that have acne and recorded their ages. Using this data, you find that the sample mean age is 18.2 years and sample standard deviation is 18.3 years. Assume that the ages are normally distributed.

To find a 91% confidence interval for the ages of people with acne, what z-value would you need to use?

Answer 1

Solution :

Given that,

We cant need to use z value.

When sample standard deviation is given then we use t- distribution.

Point estimate = sample mean = = 18.2

sample standard deviation = s = 18.3

sample size = n = 49

Degrees of freedom = df = n - 1 = 48

At 91% confidence level the t is ,

t _/2,df = t_0.045,48 = 1.730

Margin of error = E = t_/2,df * (s /n)

=  1.730* ( 18.3/ 49)

= 4.523

The 91% confidence interval estimate of the population mean is,

- E < < + E

18.2 - 4.523 < < 18.2 + 4.523

13.677 < < 22.723

(13.677 , 22.723)