This document discusses hypothesis testing using p-values. It provides an example to test whether the standard deviation of nicotine content in a sample of 100 cigarettes matches the manufacturer's claim of 2.0 mg. The sample standard deviation is calculated to be 2.4 mg. Using either the classical approach of comparing the z-score to critical values, or the p-value approach of calculating the probability of obtaining a test statistic at least as extreme as what was observed, the null hypothesis that the standard deviation equals 2.0 mg is rejected, as the evidence against it is very strong (p = 0.0046 which is less than the significance level of 0.05).

2. Test Concerning Population
VARIANCE OR STANDARD DEVIATION WHEN n ≥ 30
P- VALUE:
P- value is used in hypothesis testing to help you Accept or reject the
null hypothesis. The p value is the evidence against a null hypothesis.
The smaller the p-value, the stronger the evidence that you should
reject the null hypothesis.
MAKE DECISION BY USING P-VALUE IN Z - STATISTIC
IF P- Value ≤ α; Reject H0
IF P- Value > α; Fail to Reject H0 (Accept H0)
NOTE:
I would try to solve following examples by
1- Using classical approach
2- Using P-value approach

3. Example-1:
A random sample of 100 cigarettes of certain brand has an average
nicotine content of 20.1 mg and standard deviation of 2.4 mg. Is this
in line with the manufacturer’s claim that 𝜎 = 2.0𝑚𝑔 against 𝜎 ≠
2.0𝑚𝑔 at 5% level of significance. Assuming Normality Condition.
Solution:
n = 100, s = 2.4mg, = 0.05
1. Hypothesis
H0: 𝜎 = 2.0𝑚𝑔
H1: 𝜎 ≠ 2.0𝑚𝑔
2. Level of significance = 0.05
3. Test of statistic 𝑍 =
𝑠−𝜎0
𝜎0
√2𝑛
4. Crtical Region 𝑍∝
2
= 𝑍0.05
2
=
𝑍0.025= ±1.96
5. Computation
𝑍 =
2.4 − 2.0
2.0
√2(100)
Z = 2.83
6. Conclusion: The value of Zcal lies in the area of rejection
therefore We reject H0.


−1.96 –0– 1.96

4. MAKE DECISION BY USING P-VALUE IN Z - STATISTIC
Example-1:
A random sample of 100 cigarettes of certain brand has an average
nicotine content of 20.1 mg and standard deviation of 2.4 mg. Is this
in line with the manufacturer’s claim that 𝜎 = 2.0𝑚𝑔 against 𝜎 ≠
2.0𝑚𝑔 at 5% level of significance Assuming Normality Condition.
Solution:
n = 100, s = 2.4mg, = 0.05
1. Hypothesis
H0: 𝜎 = 2.0𝑚𝑔
H1: 𝜎 ≠ 2.0𝑚𝑔
2. Level of significance = 0.05
3. Test of statistic 𝑍 =
𝑠−𝜎0
𝜎0
√2𝑛
4. Computation
P = 2P(s ˃ 2.4)
P = 2𝑃(𝑍 ˃
2.4−2.0
2.0
√2(100)
)
P = 2P(Z ˃ 2.83)
P = 2{1- P(Z ˂ 2.83)}
P = 2(1- 0.9977) = 2(0.0023) = 0.0046
5. P-Value P-Value is 0.0046
6. Conclusion: P- Value ≤ α; Reject H0

