The article " Inconsistent Health Perceptions for US Women and Men with Diabetes" presents results of a survey of adults with diabetes. The average body mass index (BMI) in a sample of 1559 men was 30.4, with a standard deviation of 0.6. The average BMI in a sample of 1924 women was 31.1 with a standard deviation of 0.2. Find a 99% confidence interval for the difference between the mean weights.

The 99% confidence level (-0.741, -0.659).

Step-by-step explanation:

The average body mass index (BMI) in a sample of 1559 men was 30.4, with a standard deviation of 0.6.

The average BMI in a sample of 1924 women was 31.1 with a standard deviation of 0.2.

Find a 99% confidence interval for the difference between the mean weights..

The formula for 99% Confidence Interval for difference between the mean weights =

μ is the population mean, and σ is the population standard deviation.

μ1 - μ2 ± z × √σ²1/n1 + σ²2/n2

The z score for 99% confidence interval = 2.576

30.4 - 31.1 ± 2.576 × √0.6²/1559 + 0.2²/1924

-0.7 ± 2.576 × √0.36/1559 + 0.04/1924

-0.7 ± 2.576 × √0.0002309173 + 0.00002079

-0.7 ± 2.576 ×√0.0002517073

-0.7 ± 2.576 × 0.015865286

-0.7 ± 0.04086897674

Hence, the 99% Confidence Interval is

-0.7 - 0.04086897674

= -0.74086897674

≈ -0.741

-0.7 + 0.04086897674

=-0.65913102326

≈ -0.659

Therefore, the 99% confidence level (-0.741, -0.659).

only 10/60 isn't equivalent

hope this : )

none cuz there are no "following statements" regarding the question

step-by-step explanation: