Calculating P-Values for H_0: \mu =\mu_0

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SUMMARY

The discussion focuses on calculating P-values for the null hypothesis H_0: μ = μ_0 against the alternative hypothesis H_1: μ ≠ μ_0 using observed test statistics. For the test statistic z_0 = 2.45, the calculated P-value is 0.014286, derived from the equation P = 2[1 - Φ(z_0)]. In contrast, for z_0 = -0.25, the incorrect P-value of 1.19 was initially reported, which should be corrected using the relationship Φ(-z) = 1 - Φ(z) to yield the accurate P-value.

PREREQUISITES
  • Understanding of hypothesis testing and null/alternative hypotheses
  • Familiarity with normal distribution and the cumulative distribution function (CDF) Φ
  • Ability to use statistical tables for normal distributions
  • Knowledge of P-value calculation methods in statistics
NEXT STEPS
  • Study the properties of the normal distribution and its cumulative distribution function (CDF)
  • Learn about hypothesis testing frameworks and significance levels
  • Explore advanced P-value calculation techniques for different test statistics
  • Investigate common pitfalls in statistical calculations and how to avoid them
USEFUL FOR

This discussion is beneficial for students in statistics, researchers conducting hypothesis tests, and anyone involved in data analysis requiring accurate P-value calculations.

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Homework Statement


Suppose that we are testing H_0: [tex]\mu[/tex] =[tex]\mu_0[/tex] versus H_1: [tex]\mu[/tex] [tex]\neq[/tex][tex]\mu_0[/tex]. Calculate the P-value for the following observed values of the test statistic.
a: z_0 =2.45
e: z_0=-0.25


Homework Equations


none


The Attempt at a Solution


I got part a by using the equation, P=2[1-[tex]\Phi[/tex](z_0)]. I got the value for the probability from the table in the book since it's a normal distribution and the p-value was equal to .014286.
Then for part e, I tried using the same equation as before, but I got a p-value of 1.19. My equation was P=2[1-.401294]
Can someone help with what I'm doing wrong? Thanks in advance
 
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For part (e), use [itex]\Phi[/itex](-z) = 1 - [itex]\Phi[/itex](z).
 

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