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

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%)

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. (1)Ho : μ < 980

The rejection region locates on the left  tail because μ < 980

1. (2)Ho : μ > 980

The rejection region locates on the right tail because μ > 980

1. (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. (1)Ho : μ < 980

The rejection region locates on the left  tail because μ < 980

1. (2)Ho : μ > 980

The rejection region locates on the right tail because μ > 980

1. (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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