Question

The height of students in a class is distributed with mean and standard deviation. A random sample of 100 students was taken and the 90% confidence interval for mean was found to be between 175cm and 180cm.Estimate
i)value of the sample mean
Ii)value of standard deviation
Iii)95% confidence interval for mean

Answer 1

The formula for confidence interval for the mean is given by,

$\bar x\ \pm\ Z(\alpha/2)*\frac {\sigma}{\sqrt N}$

Given that,

$\mu$₁ = 180: Upper limit

$\mu$₂ = 175: Lower limit

At 90% confidence interval $\alpha = 0.1$

$\mu$₁ = 180 = $\bar x\ +\ Z(\alpha/2)*\frac {\sigma}{\sqrt N}................(Eq. \ 1)$

$\mu$2 = 175 = $\bar x\ -\ Z(\alpha/2)*\frac {\sigma}{\sqrt N}................(Eq.\ 2)$

N = 100

Hence,

$\mu1 - \mu2 = (\bar x+Z(\alpha/2)*\frac {\sigma}{\sqrt N}) - (\bar x-Z(\alpha/2)*\frac {\sigma}{\sqrt N})$

180-175 = $2*Z(\alpha/2)*\frac {\sigma}{\sqrt N}$

5 = 2 * Z_0.05*$\frac{\sigma}{\sqrt 100}$

By using the standard normal distribution table we get the Z_0.05 = 1.6449 and substituting it in the above equation we get,

$\sigma = 15.19849$

Substituting the value of mean in any one of the above equation (say eq. 1) we get,

180 = $\bar x +1.6449*\frac {15.19849}{\sqrt 100}$

Hence, $\bar x = 177.5$

ANSWER 1)

Sample mean $\bar x = 177.5$

ANSWER 2)

Standard deviation $\sigma = 15.19849$

ANSWER 3)

Using the above values for sample mean and standard deviation we can find 95% confidence interval for mean as under-

Here $\alpha = 0.05\ and\ (\alpha/2)\ = 0.025$

By using the standard normal distribution table we get the Z_0.025 = 1.96 and substituting it in the above equation 1 & 2 we get,

$\mu1 =$ 177.5 + 1.96*$\frac {15.19849}{\sqrt 100}$ = 180.4789

$\mu2 =$ 177.5 - 1.96*$\frac {15.19849}{\sqrt 100}$ = 174.521

Hence the 95% confidence interval for mean is 174.521 cms & 180.4789 cms