Question

To assess the accuracy of a laboratory scale a standard weight that is known to weigh 1 gram is repeatedly weighed 4 times. The resulting measurements (in grams) are:

0.95, 1.02, 1.01, 0.98.
For the actual weight
i) write a pivot random variable that can be used for finding confidence interval, which
distribution does this pivot have, explain your answer;
ii)find the 90% and 95% confidence intervals, state the intervals using notion of margin
of error.

Answer 1

Explanation:

how dare youuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu

Answer 2

(0.9464,1.034)

Explanation:

First find the mean μ of all values as;

(0.95+1.02+ 1.01+ 0.98 ) /4 =0.99

μ=0.99

Find the standard deviation as;

-For each value subtract the mean and square the results

0.95-0.99 = -0.04 , -0.04² = 0.0016

1.02-0.99= 0.03, 0.03²=0.0009

1.01-0.99= 0.02, 0.02²= 0.0004

0.98-0.99 = -0.01, -0.01²=0.0001

Find mean of squared deviations as;

(0.0016+0.0009+0.0004+0.0001)/4 =0.00075 ----variance

Standard deviation = √0.00075 = 0.0274

δ = 0.0274

n=4

Thus n< 30 , confidence interval for μ = x ± t*δ/√n ---------(i)

Degree of freedom (df) = n-1 =4-1 =3

The t value for 95% confidence with df= 3 is t=3.182

Substituting the values in equation (i)

0.99 ± 3.182 * 0.0274/√4

0.99 ± 3.182 * 0.0274/2

0.99 ± 3.182 * 0.0137

0.99 ± 0.0436

(0.9464,1.034)

Hope it helps...

Have a great day :P