Answer to Question #220489 in Statistics and Probability for Vickie

Question

the weight of 5 employees in a factory is as follows 50 52 49 53 50 determine the 95% confidence limit in a population mean

Answer 1

n=5

mean

$\mu = \frac{50+52+49+53+50}{5}=50.8$

standard deviation

$\sigma = \sqrt{\frac{1}{5-1}((50-50.8)^2+(52-50.8)^2+(49-50.8)^2+(53-50.8)^2+(50-50.8)^2 )} \\ = \sqrt{0.25(0.64+ 1.44+3.24+4.84+0.64 )} \\ = \sqrt{2.7} \\ = 1.643 \\ CI = (\mu - \frac{Z_c \times \sigma}{\sqrt{n}}, \mu + \frac{Z_c \times \sigma}{\sqrt{n}}) \\ Z_c=1.96 \\ CI = (50.8 - \frac{1.96 \times 1.643}{\sqrt{5}}, 50.8 + \frac{1.96 \times 1.643}{\sqrt{5}}) \\ =(50.8 -1.44, 50.8 + 1.44)$

Lower limit = 49.36

Upper limit = 52.24