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

106106​,

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

1111.

​(b) Construct

aa

9595​%

confidence interval about

muμ

if the sample​ size, n, is

2929.

​(c) Construct

aa

9999​%

confidence interval about

muμ

if the sample​ size, n, is

1111.

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

Answer 1

a)
sample mean, xbar = 106
sample standard deviation, s = 10
sample size, n = 11
degrees of freedom, n - 1 = 10

For 95% Confidence level, the t-value = 2.228
CI = (xbar - t*s/sqrt(n), xbar + t*s/sqrt(n))
CI = (106 - 2.228 * 10/sqrt(11) , 106 + 2.228 * 10/sqrt(11))
CI = (99.28 , 112.72)

b)
sample mean, xbar = 106
sample standard deviation, s = 10
sample size, n = 29
degrees of freedom, n - 1 = 28

For 95% Confidence level, the t-value = 2.048
CI = (xbar - t*s/sqrt(n), xbar + t*s/sqrt(n))
CI = (106 - 2.048 * 10/sqrt(29) , 106 + 2.048 * 10/sqrt(29))
CI = (102.2 , 109.8)

c)
sample mean, xbar = 106
sample standard deviation, s = 10
sample size, n = 11
degrees of freedom, n - 1 = 10

For 99% Confidence level, the t-value = 3.169
CI = (xbar - t*s/sqrt(n), xbar + t*s/sqrt(n))
CI = (106 - 3.169 * 10/sqrt(11) , 106 + 3.169 * 10/sqrt(11))
CI = (96.45 , 115.55)

d)
The assumption we take while calculating the CI is that the population is normally distributed.