We have the survey data on the body mass index (BMI) of 644 young women. The mea

Question

We have the survey data on the body mass index (BMI) of 644 young women. The mean BMI in the sample was bar x = 25.8. We treated these data as an SRS from a Normally distributed population with standard deviation sigma = 8.3 Give confidence intervals for the mean BMI and the margins of error for 90%, 95%, and 99% confidence. (Round your answers to two decimal places.) How does increasing the confidence level change the margin of error of a confidence interval when the sample size and population standard deviation remain the same? Increasing the confidence level causes the margin of error to decrease. Increasing the confidence level doesn't affect the margin of error. Increasing the confidence level causes the margin of error to increase. We have survey data on the body mass index (BMI) of 659 young women. The mean BMI in the sample was bar X = 28. We treated these data as an SRS from a Normally distributed population with standard deviation sigma = 7. (a) Suppose that we had an SRS of just 125 young women. What would be the margin of error for 95% confidence? (Round your answer to four decimal places)

Explanation / Answer

1. Given, mean = 25.8 Standard deviation = 8.3 n = 644

Confidence interval=

[ mean - {critical value* Standard deviation / (n^0.5)} , mean - {critical value* Standard deviation / (n^0.5)}  ]

and Margin of error = critial value* standard error

and standard error= (standard deviation) / (n^0.5) = 8.3 / (644^0.5) = 0.3270658

So, Confidence interval =

2. increase the confidence level causes the margin of error to increase

Confidence level Confidence interval margin of error Critical value 90% (25.262 to 26.34) 0.53799 1.6449 95% (25.16 to 26.44) 0.641049 1.96 99% (24.96 to 26.64) 0.842456 2.5758

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