Question

A scientist measured the speed of light. His values are in​ km/sec and have​ 299,000 subtracted from them. He reported the results of 28 trials with a mean of 756.23 and a standard deviation of 106.83.​

a) Find a 98​% confidence interval for the true speed of light from these statistics.

​b) State in words what this interval means. Keep in mind that the speed of light is a physical constant​ that, as far as we​ know, has a value that is true throughout the universe.

​c) What assumptions must you make in order to use your​ method?

​

Answer 1

a)
sample mean, xbar = 756.23
sample standard deviation, s = 106.83
sample size, n = 28
degrees of freedom, df = n - 1 = 27

Given CI level is 98%, hence α = 1 - 0.98 = 0.02
α/2 = 0.02/2 = 0.01, tc = t(α/2, df) = 2.473

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (756.23 - 2.473 * 106.83/sqrt(28) , 756.23 + 2.473 * 106.83/sqrt(28))
CI = (706.30 , 806.16)

b)
Therefore, based on the data provided, the 98% confidence interval for the population mean is 706.3 < μ < 806.16 which indicates that we are 98% confident that the true population mean μ is contained by the interval (706.3 , 806.16)

c)
the data is normally distributed
and the sample is selected randomly