Explanation:
sadly, we have only 14 data points. that is a very small sample to find a more or less reliable standard deviation for the production of softballs.
because to create the desired answer we need to calculate the mean value and the standard deviation. and we need to mix that with the Z- value of the standard normal distribution table representing the 99% confidence (= coverage of 99% of the area under the normal distribution curve, meaning therefore from 99.5% to 0.5%).
the confidence interval limits are
mean ± Z×sd/sqrt(n)
mean = mean value of the sample.
sd = standard deviation (either of - preferred - the entire population or of the sample)
n = number of data points (here 14).
Z = Z value of the standard normal distributing table for 99% (2.576).
here a little table for different confidence intervals :
Confidence Interval Z
80% 1.282
85% 1.440
90% 1.645
95% 1.960
99% 2.576
99.5% 2.807
99.9% 3.291
mean value = sum of all data points divided by the number of data points :
mean = 66.74/14 = 4.767142857...
sd = sqrt(sum((data point difference to mean)²)/n)
let's use some calculation tool like Excel (after all, the mean value is not a simple number, and to write this all up here takes a lot of space).
sd = 0.069941667...
the confidence interval limits are then
4.767142857... ± 2.576×0.069941667.../sqrt(14) =
= 4.767142857... ± 0.048152387...
upper limit = 4.815295244 ... ≈ 4.82
lower limit = 4.71899047 ... ≈ 4.72
that means, we are 99% sure that the true mean value across all produced softballs is between these 2 limits.
in other words, for 99% of such samples we can pick, their mean values will be between these 2 limits. and 1 % of such samples we expect to have a mean value outside of these 2 limits.