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HELP PLEASE!! THANKS. In a study on​ infants, one of the characteristics measured was head circumference. The mean head circumference of 14 infants was 34.9 centimeters​ (cm). The margin of error is

0.9 cm. Determine the sample size required to have a margin of error of 0.4 cm with a ​99% confidence level.
The required sample size is what. ​(Round up to the nearest whole​ number.)

User Igor Ralic
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To determine the sample size required to have a margin of error of 0.4 cm with a 99% confidence level, we can use the formula:
n = (zα/2 * σ / E)^2
where:
n = sample size
zα/2 = z-score for the given confidence level (99%), which is 2.576
σ = standard deviation of the population (unknown, but we can estimate it using the sample standard deviation)
E = margin of error, which is 0.4 cm
We are given the sample mean (34.9 cm) but not the sample standard deviation, so we need to estimate it using the margin of error.
The margin of error is defined as:
E = zα/2 * σ / sqrt(n)
Rearranging this equation, we can solve for the standard deviation:
σ = E * sqrt(n) / zα/2
Substituting the given values, we get:
0.9 cm = 2.576 * σ / sqrt(14)
σ = 0.9 cm * sqrt(14) / 2.576
σ ≈ 0.294 cm
Now we can use this estimate of the standard deviation to calculate the sample size:
n = (zα/2 * σ / E)^2
n = (2.576 * 0.294 cm / 0.4 cm)^2
n ≈ 46.7
Rounding up to the nearest whole number, we get a required sample size of 47. Therefore, we need a sample of at least 47 infants to estimate the population head circumference with a margin of error of 0.4 cm and a 99% confidence level.
User Chuk Ultima
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