Question

Bipti is a chemist at Venkatanarasimharajuvaripeta Chemical Ltd. She has published her results of the isomers of 2-Fluoropropane . The resultant masses of the isomers chosen are as follow:

5.6 gr, 5.1 gr, 4.6gr, 4.8 gr, 5.7 gr and 6.4 gr. Construct a 90 % confidence level for these sample data.

Answer 1

Table for sample mean and sample standard deviation -

Xi	(X; - XS
5.6 5.1 4.6 4.8 5.7 6.4	0.0529 0.0729 0.5929 0.3249 0.1089 1.0609
$\sum X_i = 32.2$	(X; - X)2 = 2.213

Sample size = n = 6

Sample Mean =

ΣΧ, 32.2 5.37 6

Sample standard deviation =

$S = \sqrt{\frac{\sum(X_i - \bar{X})^{2} }{n-1}} = \sqrt{\frac{2.213}{5}} = 0.665$​​​​​​90% confidence interval is given by -

$[\bar{x}\pm z_{0.10/2}*\frac{s}{\sqrt{n}}]$

where, $z_{0.10/2}$ is the Critical value of z for two tailed test at 0.10 level of significance = 1.65

(It can be obtained using the z table by finding the z corresponding to the area close to 0.05)

Hence, the confidence interval will be given by -

$[5.37\pm 1.65*\frac{0.665}{\sqrt{6}}]$

$=[5.37\pm 0.45]$

= [4.92, 5.81]