Hypothesis Tests

 

p-Value Method with small sample, σ known


Example A random sample of 10 pieces of iron wire produced in a plant of wire manufacturing company yields the mean tensile strength of sample mean X bar = 950 psi. Suppose that the population of tensile strengths of all iron wires produced in that plant are distributed with mean μ and standard deviation σ = 120 psi, test a statistical hypothesis

Ho : μ = 980

(1) Ho : μ < 980

(2) Ho : μ > 980

(3) Ho : μ ≠ 980

at the α = 0.01 level of significance. The population is assumed normal.


Suppose we have a random sample of    X1, X2, X3, ..., Xn

we have a random sample of 10 pieces of iron wire

the sample size is clearly n = 10, and in this case it is small (n < 30)


This sample is drawn from a population

The population mean is μ

The population standard deviation is σ = 120 psi


The sample mean is X bar = 950 psi

Population mean stated in the null hypothesis μ0 = 980 psi


Then we have three hypothesis we want to test about the population mean,


(1) Ho : μ < μ0

(2) Ho : μ > μ0

  1. (3)Ho : μ ≠ μ0


And, the level of significance is α = 0.01

Step 1


State the null hypothesis

Ho : μ = 980

Step 2


State the alternative hypothesis

(1) Ho : μ < 980

(2) Ho : μ > 980

(3) Ho : μ ≠ 980

Step 3


Assign an appropriate value of level of significance α

Common Values: 0.01 (1%), 0.05 (5%), and 0.10 (10%)


Given level of significance α = 0.01, in this case

Step 4


Determine a suitable test statistic

The parameter under investigation is used as test statistic


The test statistic is and call it z, n is small


z = (X bar - μ)/(σ/√n)  for  μ


This test statistic is used to test hypothesis (1), (2), (3) about μ

Step 5


Determine the probability distribution in the test statistic


Since, the sample n = 10, it is small (n < 30), according to the central limit theorem the test statistic “ z = (X bar - μ)/(σ/√n)  for  μ” is distributed as standard normal (mean 0 and standard deviation 1).

Step 6


Calculate p-value by using

Remember z is calculated value, in this case for small value of n, use this value to calculate p-value


z = (950 -980)/(120/√10)  for  μ = - 0.8


P-Value = P(Z ≤ -0.8) = 0.2119                                   if  (1) Ho : μ < μ0

[EXCEL USE NORMSDIST(-0.8)]

P-Value = P(Z ⩾ -0.8) = 1 - 0.2119 =  0.7881            if (2) Ho : μ > μ0

P-Value = 2 . P(Z ⩾ |-0.8|) = 2 * 0.2119 = 0.4237   if (3) Ho : μ ≠ μ0

Step 7


Make Decision

Compare the p-value to level of significance α, in this case 0.01


Since, p-value 0.2119 or 0.7881 or 0.4237 > α = 0.01,

we do not reject the null hypothesis for (1), (2) , and (3).

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