I just plugged in your initial question then changed the 75% to 95%. I have no clue on validity of the info but this is for 95%.

To determine the sample size needed to confirm at least 50% of the heifers are pregnant with a 95% confidence level, you can use the formula for calculating sample size for proportions in a finite population:

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 95% confidence level, Z ≈ 1.96)

P = estimated proportion (0.5 for 50%)

N = population size (25 heifers)

MOE = margin of error (let's use 0.05)

Plugging in the values:

n = [(1.96)^2 * 0.5 * (1-0.5) * 25] / [(25-1) * 0.05^2 + (1.96)^2 * 0.5 * (1-0.5)]

n = 3.8416 * 0.5 * 0.5 * 25 / (24 * 0.0025 + 3.8416 * 0.5 * 0.5)

n = 2.401 * 25 / (0.06 + 2.401)

n = 60.025 / 0.060025

n ≈ 24.996

Therefore, you would need to initially test at least 25 heifers to confirm with 95% confidence that at least 50% of the 25 heifers are pregnant. Since you cannot test a fraction of a heifer, rounding up to the nearest whole number means you should test at least 25 heifers.