Answer to Question #258395 in Statistics and Probability for princess

Question

3. Fifteen employees of a large manufacturing company were involved in a study to test a new production method. The mean production rate for the sample of 15 employees was 63 components per hour and the standard deviation was 8 components per hour. Calculate the margin of error for the mean number of components produced per hour at a 5% level of significance.

Answer 1

Lets mark the margin of error as m, then:

$m = Cr*{\frac {σ‎} {\sqrt{n}}}$ , where Cr - critical value, σ‎ - standart deviation of the sample, n - sample size

σ‎ is known, then we have to find Cr

Since we have small sample size(<30) and the population distribution is not known, it is appropiate to use t-statistics with n-1 = 14 degrees of freedom

It is also appropriate to make a two-sided test, so

$P($$t(14)>Cr) = {\frac a 2}=0.025$ , a - level os significance(0.05)

The t-values can be found from the tables

Cr = 2.145

$m= 2.145*{\frac 8 {\sqrt{15}}}=0.107=4.43$

That means that with 95% confidence the mean value lies in the interval (63-4.43, 63+4.43) =

= (58.57, 67.43)