Question

Obesity is a risk factor for many health problems such as type 2 diabetes, high blood pressure, joint problems, and gallstones. Using data collected in 2018 through the National Health and Nutrition Examination Survey, the National Institute of Diabetes and Digestive and Kidney Diseases estimates that 37.7% of all adults in the United States have a body mass index (BMI) in excess of 30 and so are categorized as obese. The data in the file Obesity are consistent with these findings.
BMI:

16.2 46.8 39 13.6 31.3
31.1 36.7 29 34.2 7.4
24.8 32.4 41 34.8 30.8
23.8 38 16 21.2 14
8 26.1 320 31 37.8
24 39.2 26 31.3 32
38.8 23.8 524 35 20.4
51.5 23.4 17 27.5 19.8
28 27.9 33 422 45.7
26.5 44.6 30 37.8 56.3
43.1 27.5 23 49.5 22.9
18.7 35.3 19 31.2
28.2 34.5 24 29
17.9 25.6 11 26.1
33.1 27.3 11 26.5
25.6 28 29 25.5

Requried:
a. Use the Obesity data set to develop a point estimate of the BMI for adults in the United States. (Round your answer to two decimal places.) Are adults in the United States obese on average?
b. What is the sample standard deviation? (Round your answer to four decimal places.)
c. Develop a 95% confidence interval for the BMI of adults in the United States. (Round your answers to two decimal places.)

Answer 1

Answer:

a. A point estimate of the BMI for adults in the United States can be calculated from the sample mean, which has a value M=44.57.

b. The sample standard deviation is s=79.9507.

c. The 95% confidence interval for the BMI of adults in the United States is (26.18, 62.96).

Step-by-step explanation:

We start by calculating the sample mean and standard deviation of the BMI data:

$M=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\M=\dfrac{1}{75}(16.2+31.1+24.8+. . .+22.9)\\\\\\M=\dfrac{3343}{75}\\\\\\M=44.57\\\\\\s=\sqrt{\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{74}((16.2-44.57)^2+(31.1-44.57)^2+(24.8-44.57)^2+. . . +(22.9-44.57)^2)}\\\\\\s=\sqrt{\dfrac{473016.8667}{74}}\\\\\\s=\sqrt{6392.1198}=79.9507$

We have to calculate a 95% confidence interval for the mean.

The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.

The sample mean is M=44.57.

The sample size is N=75.

When σ is not known, s divided by the square root of N is used as an estimate of σM:

$s_M=\dfrac{s}{\sqrt{N}}=\dfrac{79.9507}{\sqrt{75}}=\dfrac{79.9507}{8.66}=9.232$

The degrees of freedom for this sample size are:

$df=n-1=75-1=74$

The t-value for a 95% confidence interval and 74 degrees of freedom is t=1.993.

The margin of error (MOE) can be calculated as:

$MOE=t\cdot s_M=1.993 \cdot 9.232=18.39$

Then, the lower and upper bounds of the confidence interval are:

$LL=M-t \cdot s_M = 44.57-18.39=26.18\\\\UL=M+t \cdot s_M = 44.57+18.39=62.96$

The 95% confidence interval for the BMI of adults in the United States is (26.18, 62.96).