Hypothesis Tests

p-Value Method with large sample, σ known

Example A random sample of 36 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.

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

we have a random sample of    36 pieces of iron wire

the sample size is clearly n = 36, and in this case it is large (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 large

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 = 36, it is large (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 large value of n, use this value to calculate p-value

z = (950 -980)/(120/√36)  for  μ = - 1.5

P-Value = P(Z ≤ -1.5) = 0.0668                                  if  (1) Ho : μ < μ0

[EXCEL USE NORMSDIST(-1.5)]

P-Value = P(Z ⩾ -1.5) = 1 - 0.0668 =  0.9332           if (2) Ho : μ > μ0

P-Value = 2 . P(Z ⩾ |-1.5|) = 2 * 0.0668 = 0.1336   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 0.0668 or 0.9332 or 0.1336 > α = 0.01,

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

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