Question

Reserve Problems Chapter 8 Section 2 Problem 5 am 1 A random sample has been taken from a normal distribution. Output from a software package follows: Variable N Mean SE Mean StDev Variance Sum х ? ? 1.77 6.86 10 ? 762.30 3 (a) Fill in the missing quantities. in 1 ms en 2 N= Round your answers to two decimal places (e.g. 98.76). ns on 4 Mean = ms con 4 Variance = ems sion 1 (b) Find a 95% CI on the population mean. ems tion 1 Round your answers to two decimal places (e.g. 98.76). ems stion 4 Sus lems

Answer 1

a)

Now,

.6.86 SE(Mean) St Deu N N= = 15.02-15 1.77

$\\Mean = \frac{sum }{N}=\frac{762.30}{15}=50.82\\ \\ Variance=(StDev)^2=(6.86)^2=47.06$

b)

We know that the 95% confidence interval for population mean is given by:

$\\(\overline{x}\pm t_{n-1}(0.05/2)\frac{s}{\sqrt{n}})\\ \\ (50.82\pm t_{15-1}(0.05/2)\frac{6.86}{\sqrt{15}})\\ \\ (50.82\pm t_{14}(0.025)\frac{6.86}{\sqrt{15}})\\ \\ (50.82\pm [2.145]\frac{6.86}{\sqrt{15}})\\ \\ (50.82\pm 3.80)\\ \\ (47.02,54.62)$