Calculating Sample Size for Proportion Estimation

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SUMMARY

The forum discussion focuses on calculating sample sizes for proportion estimation in the context of automobile accidents involving pedestrians. For a true proportion of 0.2, the required sample size is 1179, calculated using the formula n = p(1-p) [zα/2/E]^2 with α = 0.01 and E = 0.03. In the absence of prior information about the true proportion, the sample size increases to 1842, using n = 0.25[zα/2/E]^2. The calculations were confirmed as correct based on the formulas and parameters used.

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  • Understanding of statistical concepts such as sample size determination
  • Familiarity with the normal distribution and z-scores
  • Knowledge of margin of error in proportion estimation
  • Basic arithmetic skills for calculations
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  • Learn about confidence intervals for proportions
  • Explore the impact of sample size on statistical power
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toothpaste666
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Homework Statement


  1. A national safety council wishes to estimate the proportion of automobile accidents that involve pedestrians. How large a sample of accident records must be examined to be 99% certain that the estimate does not differ from the true proportion by more than 0.03? Answer the question if

    (a) the council believes that the true proportion is 0.2.

    (b) no information about the true proportion is given.

The Attempt at a Solution


α = .01 , α/2 = .005, z.005 = 2.575
a) n = p(1-p) [zα/2/E]^2
= (.2)(.8)[2.575/.03]^2 = 1179

b) n = .25[zα/2/E]^2 = (.25)[2.575/.03]^2 = 1842

is this correct?
 
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toothpaste666 said:

Homework Statement


  1. A national safety council wishes to estimate the proportion of automobile accidents that involve pedestrians. How large a sample of accident records must be examined to be 99% certain that the estimate does not differ from the true proportion by more than 0.03? Answer the question if

    (a) the council believes that the true proportion is 0.2.

    (b) no information about the true proportion is given.

The Attempt at a Solution


α = .01 , α/2 = .005, z.005 = 2.575
a) n = p(1-p) [zα/2/E]^2
= (.2)(.8)[2.575/.03]^2 = 1179

b) n = .25[zα/2/E]^2 = (.25)[2.575/.03]^2 = 1842

is this correct?

(1) Have you used the correct formulas?
(2) Have you used the formulas correctly?
(3) Have you made no arithmetical errors?

If your answer to (1)--(3) is YES, then your answer will be correct.

You need to start having confidence in your own work. Frankly, your endless series of similar questions is starting to get old.
 
I apologize. This is the first stats class I have taken, so I wanted to make sure I understand the differences between all the different cases and that I am using the correct statistic for the correct problems. It is because I am reviewing for a final. I won't post any more hypothesis testing questions.
 
You can test the reverse direction. Given a sample of n=1179 with p=0.2, how likely is a deviation of 0.03 or more?
 

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