Question

Assuming that the population is normally distributed, construct a 90% confidence interval for the population mean for each of the samples below. Explain why these two samples produce different confidence intervals even though they have the same mean and range. Sample A: 12 3 3 6 678Full data set Sample B: 1 2 3 45678 Construct a 90% confidence interval for the population mean for sample A. (Type integers or decimals rounded to two decimal places as needed.) Construct a 90% confidence interval for the population mean for sample B. (Type integers or decimals rounded to two decimal places as needed.) Explain why these two samples produce different confidence intervals even though they have the same mean and range

Answer 1

A) Confidence interval for sample A :

From given sample data,

Sample mean = $\bar{x}$ = 4.5

Sample standard deviation = s = 2.56348

sample size = n = 8

Here population standard deviation ( $\sigma$ ) is unknown so we have to use t distribution.

Formula for confidence nterval for $\mu$ is

$CI = \bar{x} \pm t_{\alpha /2}*\frac{s}{\sqrt{n}}$

Where, $t_{\alpha /2}$ is the t critical value at given confidence level.

Here confidence level = 90% = 0.9

Significance level = $\alpha$ = 1 - 0.9 = 0.1

So t critical value at $\alpha$ = 0.1 and degrees of freedom n-1 is,

$t_{\alpha /1 , n-1} =t_{0.05,7} = 1.8946$                     

{ Using Excel, =TINV( $\alpha$ , df ) , This function returns two tailed inverse of t distribution }

So, 90% confidence nterval for $\mu$ is,

$CI = 4.5 \pm 1.8946*\frac{2.56348}{\sqrt{8}}$

$CI = 4.5 \pm 1.71713$

CI = ( 2.78 , 6.22 )

So, 90% confidence interval for population mean for sample A is,

2.78
$\leq$$\mu$$\leq$

B) Confidence interval for sample B :

From given sample data,

Sample mean = $\bar{x}$ = 4.5

Sample standard deviation = s = 2.44949

sample size = n = 8

Here population standard deviation ( $\sigma$ ) is unknown so we have to use t distribution.

Formula for confidence nterval for $\mu$ is

$CI = \bar{x} \pm t_{\alpha /2}*\frac{s}{\sqrt{n}}$

Where, $t_{\alpha /2}$ is the t critical value at given confidence level.

Here confidence level = 90% = 0.9

Significance level = $\alpha$ = 1 - 0.9 = 0.1

So t critical value at $\alpha$ = 0.1 and degrees of freedom n-1 is,

$t_{\alpha /1 , n-1} =t_{0.05,7} = 1.8946$                    

So, 90% confidence nterval for $\mu$ is,

$CI = 4.5 \pm 1.8946*\frac{2.44949}{\sqrt{8}}$

$CI = 4.5 \pm 1.6388$

CI = ( 2.86 , 6.14 )

So, 90% confidence interval for population mean for sample A is,

$\leq$$\mu$$\leq$

These two samples produce different confidence interval even though they have the same mean and range because their standard deviation are different.