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Nady [450]
3 years ago
8

The article " Inconsistent Health Perceptions for US Women and Men with Diabetes" presents results of a survey of adults with di

abetes. 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.
Mathematics
1 answer:
musickatia [10]3 years ago
5 0

Answer:

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).

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Answer:

10th term is 10

Step-by-step explanation:

The nth term for finding the geometric progression is expressed as;

Tn = ar^n-1

a is the first term

r is the common ratio

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a11 = ar^11-1

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Since a11 = -5 and r = -1/2

-5 = a(-1/2)^10

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Nest is to get the 10th terms

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