P value in population standard deviation
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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).
P value in population standard deviation
- 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