Question

We have the survey data on the body mass index (BMI) of 646 young women. The mean BMI in the sample was x̄=26.2. We treated these data as an SRS from a Normally distributed population with standard deviation σ=7.3. Find the margins of error for 95% confidence based on SRSs of N young women.

N margins of error (±0.0001)
A. 68 - 0.5901, 446 - 0.0909, 1592 - 0.0257
B. 68 - 0.2951, 446 - 0.0455, 1592 - 0.0129
C. 68 - 0.1180, 446 - 0.0182, 1592 - 0.0051
D. 68 - 0.2360, 446 - 0.0364, 1592 - 0.0103

Answer 1

To find the margins of error for 95% confidence based on SRSs of N young women, use the formula Margin of Error = Z * (standard deviation / sqrt(n)), where n is the sample size, Z is the z-score that corresponds to the chosen confidence level, and standard deviation is the population standard deviation.

Step-by-step explanation:

To calculate the margins of error for 95% confidence based on SRSs of N young women, we need to use the formula:

Margin of Error = Z * (standard deviation / sqrt(n))

where n is the sample size, standard deviation is the population standard deviation, and Z is the z-score corresponding to the desired confidence level. For a 95% confidence level, the z-score is approximately 1.96. Plugging in the values from the given information:

Margin of Error = 1.96 * (7.3 / sqrt(N))

Therefore, the margins of error for N young women are ±0.0001 * 1.96 * (7.3 / sqrt(N)).