We have the survey data on the body mass index (BMI) of 648 young women. The mean BMI in the sample was x¯=27.2. We treated these data as an SRS from a Normally distributed population with standard deviation σ= 7.4.
Find the margins of error for 90% confidence based on SRSs of N young women.
90%=1.645
N margins of error (± 0.0001)
77
353
1641
please help
To find the margin of error for a 90% confidence interval based on a simple random sample (SRS) of N young women, use the formula:
Margin of Error = Z * (σ / √N)
First, let's find the value of Z for a 90% confidence level. You mentioned that 90% corresponds to 1.645. This is incorrect. The correct value of Z for a 90% confidence level is actually 1.645.
Now, let's calculate the margin of error for each value of N:
1. For N = 77:
Margin of Error = 1.645 * (7.4 / √77) ≈ 1.858
2. For N = 353:
Margin of Error = 1.645 * (7.4 / √353) ≈ 0.743
3. For N = 1641:
Margin of Error = 1.645 * (7.4 / √1641) ≈ 0.330
Therefore, the margins of error for 90% confidence based on SRSs of N young women are approximately:
- ±1.858 for N = 77
- ±0.743 for N = 353
- ±0.330 for N = 1641