Question

A simple random sample of size n is drawn from a population that is normally distributed. The sample​ mean,

x overbarx​,

is found to be

109109​,

and the sample standard​ deviation, s, is found to be

1010.

​(a) Construct

aa

9595​%

confidence interval about

muμ

if the sample​ size, n, is

2828.

​(b) Construct

aa

9595​%

confidence interval about

muμ

if the sample​ size, n, is

1717.

​(c) Construct

anan

8080​%

confidence interval about

muμ

if the sample​ size, n, is

2828.

​(d) Could we have computed the confidence intervals in parts​ (a)-(c) if the population had not been normally​ distributed?

LOADING...

Click the icon to view the table of areas under the​ t-distribution.

Answer 1

We have given the data of random sample of size n drawn from a normally distributed population. The sample mean and sample standard deviation

(a) When n=28, calculation for 95% confidence interval for population mean

We are using the t-distribution with degrees of freedom, df=n-1

Lower limit:

Upper limit:

So, the 95% confidence interval for the population mean is (105.122, 112.878)

(b) When n=17, calculation for 95% confidence interval for population mean

Lower limit:

Upper limit:

So, the 95% confidence interval for the population mean is (103.858, 114.142)

(c) When n=28, 80% confidence interval for population mean

Lower limit:

Upper limit:

So, the 80% confidence interval for the population mean is (106.517, 111.483)

(d) The condition for the normality if the population is not normally distributed is that the sample size has to be greater than or equal to 30, i.e., .

But if the data is not normally distributed and we have small sample size then we use this formula to calculate the confidence interval for population mean which we have use to find confidence interval for part (a), (b) and (c)-   other than this there are non-parametric approach to calculate the confidence interval.