Resource assessment exercises: required sample size determination

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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 argument x and returns a numericvector containing the smallest integers not less than the corresponding elements of x E.g., 4.3 becomes 5. See also round().

  

We estimate the width of the confidence interval using the new sample size of \(n=302\).

1. 1. [1] 2.759