Resource assessment exercises: required sample size determination
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| + | [[Resource assessment exercises: standard error and confidence intervals|Previously]] we took a sample <code>S</code> of size <math>n=50</math>. | ||
| + | |||
| + | <pre> | ||
| + | S | ||
| + | |||
| + | ## [1] 22 8 18 43 21 44 17 25 32 10 11 17 9 10 56 14 14 10 20 8 37 14 55 29 33 | ||
| + | ## [26] 17 10 15 29 8 21 9 9 24 21 28 19 58 16 16 15 20 5 9 14 30 11 9 12 27 | ||
| + | </pre> | ||
| + | |||
| + | The width of the confidence interval was, | ||
| + | |||
| + | <pre> | ||
| + | sd(S)/sqrt(n) * qt(0.975, n - 1) * 2 | ||
| + | |||
| + | ## [1] 7.364 | ||
| + | </pre> | ||
| + | |||
| + | Suppose a confidence interval of <math>A=3</math> cm is desired. How large should the sample size, <math>n</math>, be? This can be estimated, | ||
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| + | {{EquationRef|equation=$A=t_{\alpha,n-1}\frac{s}{\sqrt{n} }\rightarrow n=\frac{t_{\alpha, n-1}^2s^2}{A^2}$|1}} | ||
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| + | We will use the sample <code>S</code> (<math>n=50</math>) to estimate how many observations we need in our sample. | ||
| + | In [[:wikipedia:R_(programming_language)|R]]: | ||
| + | |||
| + | <pre> | ||
| + | half.A <- 1.5 | ||
| + | req.n <- (qt(0.975, n - 1)^2 * var(S))/half.A^2 | ||
| + | req.n | ||
| + | |||
| + | ## [1] 301.2 | ||
| + | </pre> | ||
| + | |||
| + | We always need to round up! | ||
| + | |||
| + | <pre> | ||
| + | req.n <- ceiling(req.n) | ||
| + | </pre> | ||
| + | |||
| + | {{info|message=What the function <code>ceiling()</code> does|text=The function <code>ceiling()</code> ceiling takes a single numeric argument <code>x</code> and returns a numericvector containing the smallest integers not less than the corresponding elements of <code>x</code> E.g., 4.3 becomes 5. See also <code>round()</code>.}} | ||
| + | |||
| + | |||
| + | We estimate the width of the confidence interval using the new sample size of <math>n=302</math>. | ||
| + | |||
| + | |||
| + | ## [1] 2.759 | ||
[[category:Resource assessment basics in R (2014)|Required sample size determination]] | [[category:Resource assessment basics in R (2014)|Required sample size determination]] | ||
Revision as of 11:22, 23 June 2014
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This section is still under construction! This article was last modified on 06/23/2014. If you have comments please use the Discussion page or contribute to the article! |
Previously we took a sample S of size \(n=50\).
S ## [1] 22 8 18 43 21 44 17 25 32 10 11 17 9 10 56 14 14 10 20 8 37 14 55 29 33 ## [26] 17 10 15 29 8 21 9 9 24 21 28 19 58 16 16 15 20 5 9 14 30 11 9 12 27
The width of the confidence interval was,
sd(S)/sqrt(n) * qt(0.975, n - 1) * 2 ## [1] 7.364
Suppose a confidence interval of \(A=3\) cm is desired. How large should the sample size, \(n\), be? This can be estimated,
| $A=t_{\alpha,n-1}\frac{s}{\sqrt{n} }\rightarrow n=\frac{t_{\alpha, n-1}^2s^2}{A^2}$ | 1 |
We will use the sample S (\(n=50\)) to estimate how many observations we need in our sample.
In R:
half.A <- 1.5 req.n <- (qt(0.975, n - 1)^2 * var(S))/half.A^2 req.n ## [1] 301.2
We always need to round up!
req.n <- ceiling(req.n)
What the function ceiling()does- The function
ceiling()ceiling takes a single numeric argumentxand returns a numericvector containing the smallest integers not less than the corresponding elements ofxE.g., 4.3 becomes 5. See alsoround().
We estimate the width of the confidence interval using the new sample size of \(n=302\).
- [1] 2.759
