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One-Sample Z-Test: Hypothesis Testing for Population Mean, Study notes of Experimental Design

College of EuropeExperimental Design

The one-sample z-test, a statistical hypothesis testing method used to determine if the population mean of a single group is significantly different from a hypothesized constant. Assumptions, inputs, five steps for performing the test, and a discussion on the distribution of the test statistic. A sample problem is provided for illustration.

Typology: Study notes

2021/2022

Uploaded on 08/05/2022

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One-sample z-test
1. Assumptions
•Experimental Design: The sample forms a single treatment group.
•Null Hypothesis: The population mean of the treatment group is
not significantly different from a hypothesized constant c.
•Population Distribution: Arbitrary.
•Sample Size: Greater than or equal to 30.
2. Inputs for the z-test
•Sample size: n
•Sample mean: ¯x
•Sample standard deviation: sx
•Standard error of mean: SEmean =sx
√n
•Null hypothesis value: c
•The level of the test: α
3. Five Steps for Performing the Test of Hypothesis
1. State null and alternative hypotheses:
H0:µ=c, H1:µ6=c
2. Compute test statistic:
z=¯x−µ
SEmean
,
assuming the null hypothesis value for µ.
3. Compute 100(1 −α)% confidence interval Ifor z.
4. If z∈I, accept H0; if z /∈I, reject H1and accept H0.
5. Compute p-value.
1
pf3

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One-sample z-test

1. Assumptions

  • Experimental Design: The sample forms a single treatment group.
  • Null Hypothesis: The population mean of the treatment group is not significantly different from a hypothesized constant c.
  • Population Distribution: Arbitrary.
  • Sample Size: Greater than or equal to 30.

2. Inputs for the z-test

  • Sample size: n
  • Sample mean: ¯x
  • Sample standard deviation: sx
  • Standard error of mean: SEmean =

sx √ n

  • Null hypothesis value: c
  • The level of the test: α

3. Five Steps for Performing the Test of Hypothesis

  1. State null and alternative hypotheses:

H 0 : μ = c, H 1 : μ 6 = c

  1. Compute test statistic: z =

x¯ − μ SEmean

assuming the null hypothesis value for μ.

  1. Compute 100(1 − α)% confidence interval I for z.
  2. If z ∈ I, accept H 0 ; if z /∈ I, reject H 1 and accept H 0.
  3. Compute p-value.

4. Discussion

Since we are assuming that n ≥ 30, the sample standard deviation sx is a close approximation to the population standard deviation σx; we will assume that σx is known and equal to sx. Furthermore, since SEmean is constant, E(¯x) = μx, and SEmean = sx/

n = σx/

n,

E(z) = E

x¯ − μx SEmean

E(¯x) − μx SEmean

μx − μx SEmean

and

Var(z) = Var

x¯ − μx SEmean

Var(¯x − μx) SE^2 mean

Var(¯x) SE^2 mean

SE^2 mean SE^2 mean

Thus, by the central limit theorem, z has an approximately standard normal distribution and we can use the standard normal table to compute confidence intervals and p-values for z.

4. A Sample Problem

Mendenhall and Sincich, p. 45: Humerous bones from the same species of animal have approximately the same length-to-width ratios. It is known that Species A has a mean ratio of 8.5. Suppose that 41 fossil humerous bones were unearthed at a site where Species A is known to have flourished. (We assume that all bones are from the same species.) The length-to-width ratios of these bones has sample mean 9.26 and sample standard deviation 1.20. Can we conclude that these bones belong to Species A? Perform a level 0. z-test to check.

Solution: We have these inputs:

n = 41 ¯x = 9. 26 sx = 1. 20 c = 8. 5 α = 0. 05

and compute

SEmean =

σx √ n

Here are the five steps of the z-test:

  1. State the null and alternative hypotheses:

H 0 = 8. 5 , H 1 6 = 8. 5