Question

Suppose we want to estimate the average weight of an adult male in Dekalb County, Georgia. We draw a random sample of 1,000 men from the population and weigh them. We find that the average man in our sample weighs 190 pounds, and the standard deviation of the sample is 30 pounds. What is the 95% confidence interval?

Answer 1

Answer: (188.1384, 191.8616 )

Explanation:

When population standard deviation is not known, then the confidence interval for population mean is given by :-

`\overline{x}\pm t^* (s)/(√(n))` (1)

, where n= sample size.

`\overline{x}` = sample mean

t* = critical value.

s = sample standard deviation.

As per given , we have

n= 1000

Degree of freedom : df = n-1=999

`\overline{x}=190`

s= 30

Significance level :
`\alpha=1-0.95=0.05`

Using t-distribution table , the critical value =
`t^*=t_((\alpha/2,df))=t_((0.025, 999))=1.9623`

The 95% confidence interval will be :

`190\pm (1.9623) (30)/(√(1000))`

`190\pm (1.9623) (0.948683298051)`

`190\pm 1.86160123577`

`\approx(190- 1.8616,\ 190+ 1.8616)=(188.1384,\ 191.8616 )`

Hence, the required confidence interval = (188.1384, 191.8616 )