Question

For the following data: 2 3 5 6.6 7.1 7.8 4.7 8.5 5.4 a. Calculate the sample mean and sample variance b. Calculate the probability that the population mean is between 9 and 10 if the population standard deviation is known to be 1.5. c. What is the 98% confidence interval for the population mean if the population standard deviation is known to be 1.5? d. Calculate the 98% confidence interval using the sample standard deviation.

Answer 1

a) Sample mean =   = 40.1 / 6 = 6.6833

Σ. .. n

Sample variance =   =  2.0617

Ꮖ -n*.7) n-1

b) Assuming that the data follows a normal distribution,

X
  $\sim$ N( $\mu$ = 6.6833,  $\sigma^{2}/n$ = 0.375)
P( 9 <
X
< 10 ) =  $P(\frac{9-6.6833}{\sqrt{0.375}}<\frac{\bar{X}-\mu}{\sigma^{2}/n}<\frac{10-6.6833}{\sqrt{0.375}})$
= P( 3.7832 < Z < 5.416 )
= P( Z < 5.416 ) - P( Z < 3.7832 )
= 1 - 0.9999
= 0.0001

c) In general, (1-$\alpha$)% confidence interval for population mean is obtained as follows :
$(\bar{x}\mp Z_{\alpha /2}*\sigma*\sqrt{\frac{1}{n}})$
Here, Z ($\alpha$/2) = Z (0.01) = 2.3263

Thus, 98% confidence interval for population mean is :
$(6.6833-2.3263*1.5*\sqrt{\frac{1}{6}},6.6833+2.3263*1.5*\sqrt{\frac{1}{6}})$
$\Rightarrow$ ( 5.2587, 8.1079 )

d) In general, (1-$\alpha$)% confidence interval for population mean is obtained as follows :
$(\bar{x}\mp t_{\alpha /2,n-1}*s*\sqrt{\frac{1}{n}})$
Here, t ($\alpha$/2, n-1) = t (0.01, 5) = 3.3648   & s = sqrt(2.0617) = 1.4359

Thus, 98% confidence interval for population mean is :
$(6.6833-3.3648*1.4359*\sqrt{\frac{1}{6}},6.6833+3.3648*1.4359*\sqrt{\frac{1}{6}})$
$\Rightarrow$ ( 4.7108, 8.6558 )

Hope this answers your query!