🧑🏼‍🔬 Z Test

Standard Normal Deviate Tests or Large Sample Tests

• If the sample size n > 30 then it is considered as large sample and if the sample size n < 30 then it is considered as small sample.

SND Test or One Sample (Z-test)

• Given by R.A. Fisher.

Case-I: Population standard deviation (σ) is known Assumptions:

Assumptions:

• Population is normally distributed
• The sample is drawn at random

Conditions:

• Population standard deviation σ is known
• Size of the sample is large (say n > 30)

Procedure:

• Let x₁, x₂, ……… x_n be a random sample size of n from a normal population with mean μ and variance σ².
• Let x be the sample mean of sample of size “n”
• Null hypothesis (H₀): population mean (µ) is equal to a specified value μ₀
• i.e. H₀ : µ = µ₀
• Under H₀, the test statistic is

• i.e. the above statistic follows Normal Distribution with mean “0” and varaince ‟1”. If the calculated value of Z < table value of Z at 5% level of significance, H₀ is accepted and hence we conclude that there is no significant difference between the population mean and the one specified in H₀ as µ₀.

Case-II: If σ is not known

Assumptions:

• Population is normally distributed
• Sample is drawn from the population should be random
• We should know the population mean Conditions:
• Population standard deviation σ is not known
• Size of the sample is large (say n > 30)
• Null hypothesis (H₀): µ = µ₀ under H₀, the test statistic

• If the calculated value of Z < table value of Z at 5% level of significance, H₀ is accepted and hence we conclude that there is no significant difference between the population mean and the one specified in H₀ otherwise we do not accept H₀.
• The table value of Z at 5% level of significance = 1.96 and table value of Z at 1% level of significance = 2.58.

Two sample Z-Test or Test of significant for difference of means

Case-I: When σ is known

Assumptions:

• Populations are distributed normally
• Samples are drawn independently and at random Conditions:
• Populations standard deviation σ is known
• Size of samples are large

Procedure:

• Let x₁ be the mean of a random sample of size n₁ from a population with mean µ₁ and variance σ₂²
• Let x₂ be the mean of a random sample of size n₂ from another population with mean µ₂ and variance σ₂²
• Null hypothesis H₀: σ₁ = σ₂
• Alternative Hypothesis H₁: σ₁ ≠ σ₂
• i.e. The null hypothesis states that the population means of the two samples are identical. Under the null hypothesis the test statistic becomes

• i.e. the above statistic follows Normal Distribution with mean “0” and variance “1”. If σ₁² = σ₂² = σ² (say) i.e. both samples have the same standard deviation then the test statistic becomes

• If the calculated value of |Z| < table value of Z at 5% level of significance, H₀ is accepted.
• Otherwise rejected. If H₀ is accepted means, there is no significant difference between two population means of the two samples are identical.

Example: The Average panicle length of 60 paddy plants in field No. 1 is 18.5 cms and that of 70 paddy plants in field No. 2 is 20.3 cms. With common S.D. 1.15 cms. Test whether there is significant difference between two paddy fields w.r.t panicle length.

Solution:

• Null hypothesis: H₀: There is no significant difference between the means of two paddy fields w.r.t. panicle length.
• H₀: μ₁ = μ₂
• Under H₀, the test statistic becomes

• Substitute the given values in equation (1), we get

  

• Calculated value of Z = 8.89
• Calculated Value of Z > table value of Z at 5% LOS (1.96), H₀ is rejected. This means, there is highly significant difference between two paddy fields w.r.t. panicle length.

Case-II: When σ is not known

Assumptions:

• Populations are normally distributed
• Samples are drawn independently and at random Conditions:
• Population standard deviation σ is not known
• Size of samples are large

Null hypothesis H₀: μ₁ = μ₂

• Under the null hypothesis the test statistic becomes

• If the calculated value of |Z| < table value of Z at 5% level of significance, H₀ is accepted otherwise rejected.

Example

• A breeder claims that the number of filled grains per panicle is more in a new variety of paddy ACM.5 compared to that of an old variety ADT.36. To verify his claim a random sample of 50 plants of ACM.5 and 60 plants of ADT.36 were selected from the experimental fields.
• The following results were obtained:

  

Sol:

• Null hypothesis H₀: μ₁ = μ₂ (i.e. the average number of filled grains per panicle is the same for both ACM.5 and ADT.36)
• Under H₀, the test statistic becomes

• Substitute the given values in equation (1), we get

  

• Calculated value of Z > Table value of Z at 5% LOS (1.96), H₀ is rejected. We conclude that the number of filled grains per panicle is significantly greater in ACM.5 than in ADT.36