The article "uncertainty estimation in railway track life-cycle cost"† presented the following data on time to repair (min) a rail break in the high rail on a curved track of a certain railway line.
159 120 480 149 270 547 340 43 228 202 240 218
a normal probability plot of the data shows a reasonably linear pattern, so it is plausible that the population distribution of repair time is at least approximately normal. the sample mean and standard deviation are 249.7 and 145.1, respectively.
(a) is there compelling evidence for concluding that true average repair time exceeds 200 min? carry out a test of hypotheses using a significance level of 0.05.
state the appropriate hypotheses.
h0: μ > 200
ha: μ = 200h0: μ < 200
ha: μ = 200 h0: μ = 200
ha: μ ≠ 200h0: μ = 200
ha: μ > 200h0: μ = 200
ha: μ < 200
calculate the test statistic and determine the p-value. (round your test statistic to two decimal places and your p-value to four decimal places.)
t =
p-value =
what can you conclude?
there is compelling evidence that the true average repair time exceeds 200 min.
there is not compelling evidence that the true average repair time exceeds 200 min.
(b) using
σ = 150,
what is the type ii error probability of the test used in (a) when true average repair time is actually 300 min? that is, what is β(300)? (round your answer to two decimal places.)
β(300) =

step-by-step explanation:

i found the life tables for the year 2012. 

life table of males at age 30. probability of dying between ages x and x+1

total population:  

29–30 . . . . . . . . . . . . 0.001030

30–31 . . . . . . . . . . . . 0.001061

males:  

29–30 . . . . . . . . . . . . 0.001420

30–31 . . . . . . . . . . . . 0.001448

white population

29–30 . . . . . . . . . . . . 0.001005

30–31 . . . . . . . . . . . . 0.001037

white males

29–30 . . . . . . . . . . . . 0.001372

30–31 . . . . . . . . . . . . 0.001399

non hispanic white population

29–30 . . . . . . . . . . . . 0.001083

30–31 . . . . . . . . . . . . 0.001116

non hispanic white males

29–30 . . . . . . . . . . . . 0.001477

30–31 . . . . . . . . . . . . 0.001503

jacob, non hispanic white male. 30 years old.

probability that jacob lives to be 30 years old. 

since the given probabilities are probabilities of dying, i think we simply subtract it from 1 to get the probability of living. we will use the probability of non hispanic white male.

                                          probability of living to be 30 y.o.

29 - 30 : 1 - 0.001477 = 0.998523

30 - 31 : 1 - 0.001503 = 0.998497

you didn't post any answer choices, so its not 100% clear what your teacher is looking for

however, log(15/7) is equivalent to log(15) - log(7)

i'm using the rule log(x/y) = log(x) - log(y)