Question

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
H_a: μ = 200H₀: μ < 200
H_a: μ = 200    H₀: μ = 200
H_a: μ ≠ 200H₀: μ = 200
H_a: μ > 200H₀: μ = 200
H_a: 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) =

Answer 1

Concepts and reason

The t-distribution is the continuous distribution where the sample size is small and population standard deviation is unknown.

If the test does not match with the reality then error occurred. There are 2 types of errors,

Type I and type II error.

Fundamentals

Null hypothesis states that there is no statistical significance exists in given set of observations, Type I error is rejecting ${H_0}$ when ${H_0}$ is actually true. The probability of type I error is represented by $\alpha$ .

Alternative hypothesis states that there is statistical significance exist in given set of observations. Type II error is accepting ${H_0}$ when it is not true. The probability of type II error is represented by $\beta$ .

The probability value or p -value is the smallest value at which ${H_0}$ can be rejected. It is used to determine the results of the hypothesis that is whether to reject or accept ${H_0}$ . Power of the test is the probability of rejecting ${H_0}$ when it is false. It is provided by $1 - \beta$ .

The test statistic of t-distribution is defined as,

${\rm{ }}t = \frac{{\bar X - \mu }}{{S/\sqrt n }}$

Where the sample mean is $\left( {\bar X} \right)$ , $S$ is sample standard deviation, $n$ is the sample size and $\mu$ is the population mean. The p-value can be obtained by using excel function

$= TDIST\left( {} \right)$

Power of the test is defined as follows,

$1 - \beta = P\left( {Z > \left( {{z_\alpha } - \frac{{{\mu _1} - {\mu _0}}}{{\sigma /\sqrt n }}} \right)} \right)$

Where, ${z_\alpha }$ is the critical value, $\sigma$ is the population standard deviation, ${\mu _1}$ and ${\mu _0}$ are the two means.

(a.1)

The hypothesis to test the claim that true average repair time exceeds 200 is defined as,

The null and the alternate hypothesis can be stated as follows,

$\begin{array}{l}\\{H_0}:\mu = 200\\\\{H_1}:\mu > 200\\\end{array}$

So, the hypothesis ${H_a}:\mu = 200,{H_0}:\mu < 200$ , ${H_a}:\mu = 200,{H_0}:\mu = 200$ is incorrect as in alternative hypothesis true average repair time is equal to 200 which is not possible.

The hypothesis ${H_a}:\mu \ne 200,{H_0}:\mu = 200$ is incorrect as the hypothesis is defined for two tail tests and to claim that true average repair time exceeds 200 is one tail test.

The hypothesis ${H_a}:\mu < 200,{H_0}:\mu > 200$ is incorrect as here the null hypothesis is incorrectly defined.

The hypothesis to test the claim that true average repair time exceeds 200 is defined as,

The null and the alternate hypothesis can be stated as follows,

$\begin{array}{l}\\{H_0}:\mu = 200\\\\{H_a}:\mu > 200\\\end{array}$

(a.2)

The population distribution of repair time is approximately normal with sample mean as 249.7 and standard deviation as 145.1. The sample size is 12. So, the test statistic value of t-test is calculated as,

$\begin{array}{c}\\t = \frac{{\bar X - \mu }}{{S/\sqrt n }}\\\\ = \frac{{249.7 - 200}}{{145.1/\sqrt {12} }}\\\\ = 1.19\\\end{array}$

The p-value is calculated using excel.

The p-value is 0.1295.

(b)

From the provided information,

$\begin{array}{l}\\{\mu _1} = 300\\\\{\mu _0} = 200\\\\\sigma = 150\\\\n = 12\\\end{array}$

The ${z_\alpha }$ can be calculated using $NORMSINV\left( {} \right)$ function of excel as follows,

$\begin{array}{c}\\1 - \beta = P\left( {Z > \left( {{z_\alpha } - \frac{{{\mu _1} - {\mu _0}}}{{\sigma /\sqrt n }}} \right)} \right)\\\\ = P\left( {Z > 1.645 - \frac{{300 - 200}}{{150/\sqrt {12} }}} \right)\\\\ = P\left( {Z > - 0.6644} \right)\\\\ = P\left( {Z < 0.6644} \right)\\\end{array}$

The $P\left( {Z < 0.6644} \right)$ is calculated using the $NORMSDIST\left( {} \right)$ function of excel as follows,

Then,

$\begin{array}{c}\\1 - \beta = 0.75\\\\\beta = 1 - 0.75\\\\\beta = 0.25\\\end{array}$

Ans: Part a.1

The hypothesis to claim that true average repair time is,

$\begin{array}{l}\\{H_0}:\mu = 200\\\\{H_a}:\mu > 200\\\end{array}$

Part a.2

The p-value and test statistic value is,

t	=	$1.19$
P-value	=	$0.1295$

Part b

The value of probability of type II error or $\beta$ is $0.25$ .