Hypothesis Tests
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
(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).
DNA Pot (c) 2009 - Current