Inferential statistics means making inferences about the population using the sample data. To know weather the assumption about the population parameter is true or not Hypothesis testing is used.
Hypothesis testing is used to confirm your conclusion (or hypothesis) about the population parameter .Through hypothesis testing, you can determine whether there is enough evidence to conclude if the hypothesis about the population parameter is true or not.
Hypothesis Testing starts with the formulation of these two hypotheses:
- Null hypothesis (H₀): The status quo.
- The alternate hypothesis (H₁): The challenge to the status quo.
You can use the following rule to formulate the null and alternate hypotheses:
The null hypothesis always has the following signs: = OR ≤ OR ≥
The alternate hypothesis always has the following signs: ≠ OR > OR <
To summarize this, you cannot decide the status quo or formulate the null hypotheses from the claim statement, you need to take care of signs in writing the null hypothesis. Null Hypothesis never contains ≠ or > or < signs. It always has to be formulated using = or ≤ or ≥ signs.
The formulation of the null and alternate hypotheses determines the type of the test and the position of the critical regions in the normal distribution.
You can tell the type of the test and the position of the critical region based on the ‘sign’ in the alternate hypothesis.
≠ in H₁ → Two-tailed test → Rejection region on both sides of the distribution.
< in H₁ → Lower-tailed test → Rejection region on the left side of the distribution.
> in H₁ → Upper-tailed test → Rejection region on right side of the distribution.
Critical region gives the statistical reason to reject the null hypothesis.
Critical Value Method :
Critical values are and how your decision to reject or fail to reject the null hypothesis (we never accept the null hypothesis) is based on the critical values and the position of the sample mean on the distribution.
Example — A nerologist is testing the effect of drug on response time by injecting 100 rats with a unit dose of the drug,subjecting each to neurological stimulus and recording its response time. The nerologist knows that the mean response time for rats not injected with the drug is 1.2 seconds. The mean of the 100 injected rats response times is 1.05 seconds with a sample standard deviation of 0.5 seconds. Do you think that the drug has effect on the response time.
Solution :
Null Hypothesis (H₀) — Drug has no effect i.e μ₀ = 1.2 seconds.
Alternate Hypothesis (H₁) — Drug has effect ie μ₀≠ 1.2 seconds.
Assume H₀ is true, now in the sampling distribution sample mean μₓ = μ₀ = 1.2 seconds and σ ₓ= σ / √n ≈ S/√n= 0.5 /√100 = 0.05 ( population mean is assumed as equal to sample mean ).
What is the probability of getting 1.05 sec ( or how many standard deviations is 1.05 sec away from sample mean of 1.2 sec ).
Z = (x -μₓ)/ σ ₓ
Z = (1.05–1.2)/0.05 = 3
α -> significance level for the test. It refers to the proportion of the sample mean lying in the critical region.This can be also stated as probability of making an error.
For this test, α is taken as 0.05 (or 5%).
Since critical region lies between both sides , the acceptance region is 95 % and rejection region is 5 % , this 5% is split on both sides hence on each side it is 2.5 % .
Zc (for 2.5 % probability (calculated from z table))= +1.96 or -1.96.
Since Z for mean of 1.05 sec is 3 which lies in critical region we reject the null hypothesis and say that the drugs has the effect on the response time.
Alternate way,
UCV — Upper critical value — μ + (Zc * σ ₓ) = 1.2 + 1.96*0.05 = 1.298
LCV — Lower critical value — μ -(Zc * σ ₓ) = 1.2 -1.96*0.05 = 1.102
Since 1.05 sec is below 1.102 (LCV) lies in the critical region we reject the null hypothesis.Hope the concept was clear.
In short, steps to perform critical value method:
Thanks for reading !!
