Question

Write an expression (involving μ and σ) for a tolerance interval containing 98 percent of all the population measurements.

Answer 1

t-distribution tables tell you that a 98% confidence interval ranges from the (sample mean - 2.327 standard deviations) to (the sample mean + 2.327 standard deviations).

\[xbar - 2.327*(\Sigma/\sqrt{n}) < \mu < xbar + 2.327*(\Sigma/\sqrt{n})\]