Hypothesis Tests
Hypothesis Tests
Test population Mean with Small sample size, , 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.Assume population is normal.
From the question:
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 assumed normal.
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%)
In this case it is given: α = 0.01
Step 4
Determine a suitable test statistic
The parameter under investigation is used as test statistic
The test statistic is
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 is small n = 10 < 30, according to the question, since it is assumed normal, “z = (X bar - μ)/(σ/√n) for μ” is distributed as standard normal (mean 0 and standard deviation 1). Or if it is not assumed normal, do normal plot and boxplot to check normality.
Step 6
Locate all rejection region and find the critical point
The location of rejection region depends on the alternative hypothesis.
(1)Ho : μ < 980
The rejection region locates on the left tail because μ < 980
(2)Ho : μ > 980
The rejection region locates on the right tail because μ > 980
(3)Ho : μ ≠ 980
The rejection region locates on both tails because μ ≠ 980
Step 7
Find the critical point
The critical point depends on the value assigned to size of the rejection region or the level of significance which is type I error probability α
the level of significance = P( type I error) = α = 0.01
the critical point values can be read out of the table for α = 0.01
in each case (1),(2), and (3)
Level of significance α Left-tail Right Tail Both Tail
0.01 -2.33 +2.33 -2.575 and +2.575
0.05 -1.645 +1.645 -1.96 and +1.96
0.10 -1.28 +1.28 -1.645 and +1.645
location of rejection region depends on the alternative hypothesis.
(1)Ho : μ < 980
The rejection region locates on the left tail because μ < 980
(2)Ho : μ > 980
The rejection region locates on the right tail because μ > 980
(3)Ho : μ ≠ 980
The rejection region locates on both tails because μ ≠ 980
Step 8
Calculate the test statistic
Take a random sample from a population in question, calculate the value of the test statistic.
The random the sample is 36
And n=10 is n < 30
Calculate X bar = 950 in this case given
σ = 120 and n =10
μ is the μ0 = 950 from the null hypothesis
use all values in calculating test statistic “ z = (X bar - μ)/(σ/√n) for μ”
z = (X bar - μ)/(σ/√n) for μ = (950 - 980) / (120/√10) = -0.8
Step 9
Make decision
Determine whether it falls in the rejection region. If it falls in the rejection region, we reject the null hypothesis. If it doesn’t fall in the rejection region, keep it.
Check the value from the calculation of test statistic
z = (X bar - μ)/(σ/√n) for μ = -0.8
This value -1.5 is checked whether or not falls into one of the rejection regions. Found out that -2.33 < -0.8 <+2.33 and -2.575<-0.8<+2.575.
Therefore, -0.8 does not fall
in these rejection regions.
The decision is to keep the null
hypothesis Ho : μ = 980.
The data supports that the mean tensile strength of the iron wire is
980 psi.
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