How to Determine Whether to Accept or Reject a Hypothesis?

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To determine whether to accept or reject the null hypothesis in hypothesis testing, it's essential to calculate the mean (μ) and standard deviation (σ) based on the sample size and number of successes. The confusion arises around using continuity correction when calculating probabilities, specifically whether to use values like 22.5 or 23.5. The discussion highlights that while integer values are significant in a binomial distribution, approximating with a normal distribution can simplify calculations. The importance of continuity correction is emphasized, as it helps in accurately assessing probabilities in discrete distributions. Understanding these concepts is crucial for making informed decisions in hypothesis testing.
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Summary:: Confused on correction for contunuity

I'm confused about one step for questions like this where you have to determine to accept or reject the null hypothesis.
Sample size: 50
Number of successes: 23
Significance level: 10%
Null hypothesis: p = 0.4
Accepted hypothesis: p > 0.4

Solution:
μ = np
= (50)(0.4)
= 20

σ = √npq
= √(20)(0.6)
= 3.46

This is the part I'm confused on: To determine if I should accept/reject the hypothesis, would I do 1) or 2) or 3)

1) P ( x̅ >23)
= P ( x̅ > 23.5)
= 1-P(x̅<23.5)
and then solve

2) P ( x̅ >23)
= 1 - ( x̅ < 22.5)
and then solve

3) P ( x̅ < 23)
=P ( x̅ < 22.5)
and then solve
 
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Why do you have to introduce 22.5 and 23.5 at all? Can't you just look at ##P(\bar{x} \geq 23)##?

What is your plan with the ##\sigma## you computed? Typically that means you are going to pretend that the values you are drawing are from some t distribution or something, and the fact that it's integer valued no longer is as much of an issue, since in the approximation the integers are not special values.
 
Office_Shredder said:
Why do you have to introduce 22.5 and 23.5 at all? Can't you just look at ##P(\bar{x} \geq 23)##?

What is your plan with the ##\sigma## you computed? Typically that means you are going to pretend that the values you are drawing are from some t distribution or something, and the fact that it's integer valued no longer is as much of an issue, since in the approximation the integers are not special values.
I would guess that in the question the samples are drawn from a binomial distribution and so the fact that it is integer valued is important. This question belongs in the homework section, I'll get it moved.
 

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