This is what the AI just spit out.
To determine the sample size needed to confirm at least 50% of the heifers are pregnant with a 75% confidence level, you can use a formula for calculating sample size for proportions in a finite population.
The formula is:
n = [Z^2 * P(1-P) * N] / [(N-1) * MOE^2 + Z^2 * P(1-P)]
where:
n = sample size
Z = Z-score corresponding to the desired confidence level (for 75% confidence level, Z ≈ 1.15)
P = estimated proportion (0.5 for 50%)
N = population size (25 heifers)
MOE = margin of error
Assuming a margin of error of 0.05 (5%), you can plug in the values:
n = [(1.15)^2 * 0.5 (1-0.5) * 25] / [(25-1) * 0.05^2 + (1.15)^2 * 0.5 (1-0.5)]
n = 2.56 / (24 * 0.0025 + 1.56)
n = 2.56 / (0.06 + 1.56)
n ≈ 2.56 / 1.62
n ≈ 1.58
Therefore, you would need to initially test at least 2 heifers to confirm with 75% confidence that at least 50% of the 25 heifers are pregnant.
Flemingfarm
npub1sz…qn3du
2024-07-07 17:50:02
in reply to nevent1q…u5qn
Author Public Key
npub1szlawrnh4kzvq9yfuegvpz0ws8h5zywea4n2d4zf3kdd3g9tenwqnqn3duPublished at
2024-07-07 17:50:02Event JSON
{
"id": "9b370603031a95e57c428710030173c61a157b4b48d70481f0734486f828ad05",
"pubkey": "80bfd70e77ad84c01489e650c089ee81ef4111d9ed66a6d4498d9ad8a0abccdc",
"created_at": 1720367402,
"kind": 1,
"tags": [
[
"e",
"3666e65ededfb9dc367dddc4574b2a46b4e6d6262c9106c7499b5f7e27c56d5c",
"",
"root"
],
[
"p",
"d70d50091504b992d1838822af245d5f6b3a16b82d917acb7924cef61ed4acee"
]
],
"content": "This is what the AI just spit out.\n\nTo determine the sample size needed to confirm at least 50% of the heifers are pregnant with a 75% confidence level, you can use a formula for calculating sample size for proportions in a finite population.\n\nThe formula is:\n\nn = [Z^2 * P(1-P) * N] / [(N-1) * MOE^2 + Z^2 * P(1-P)]\n\nwhere:\nn = sample size\nZ = Z-score corresponding to the desired confidence level (for 75% confidence level, Z ≈ 1.15)\nP = estimated proportion (0.5 for 50%)\nN = population size (25 heifers)\nMOE = margin of error\n\nAssuming a margin of error of 0.05 (5%), you can plug in the values:\n\nn = [(1.15)^2 * 0.5 (1-0.5) * 25] / [(25-1) * 0.05^2 + (1.15)^2 * 0.5 (1-0.5)]\nn = 2.56 / (24 * 0.0025 + 1.56)\nn = 2.56 / (0.06 + 1.56)\nn ≈ 2.56 / 1.62\nn ≈ 1.58\n\nTherefore, you would need to initially test at least 2 heifers to confirm with 75% confidence that at least 50% of the 25 heifers are pregnant.",
"sig": "51120b5a8b505874c7781aba6a0930bc9101781f46d0d422c9aee9a90097cc8c095396ce9fbf2d8d73a46798d3b9f0c052cb4da283bfa1489f8b6c46c4de1d38"
}