Question

Question 4: To access the accuracy of a laboratory scale, a standard weight known to weigh 10g is weighed repeatedly. The scale readings are normally distributed with standard t. The weight is measured five times. The mean result is 0.0023g. Give a 91.64% b. How rnany measurements needed to get the bound on the error of estimation with 98% deviation 0.0002g confidence interval for the mean of the true weight. confidence is less than 0.0001?

Answer 1

SolutionA:

sample mean=10.0023g

n=5

alpha=1-0.9164=0.0836

given std deviation=0.0002 g

91.64 % confidence interval for mean is given by

xbar-margin of error and xbar+margin of error

margin of error in excel is

=CONFIDENCE.NORM(0.0836;0.0002;5)

=0.000155

91.64 % confidence interval for mean of true weight is given by

10.0023-0.000155 and 10.0023+0.000155

10.00215 and 10.00245

lower limit= 10.00215

upper limit=10.00245

Solutionb:

E=0.0001

z crit for 98%=2.326

n=(Z alpha/2*sigma/E)^2

=(2.326*0.0002/0.0001)^2

n=21.64

n=22

Required sample szie=n=22