Question

In a test of the Atkins weight loss program, 40 individuals participated in a randomized trial with overweight adults. After 12 months, the mean weight loss was found to be 2.1 lb, with a sample standard deviation of 4.8 lb.

a. What is the best point estimate of the population mean weight loss of all overweight adults who follow the Atkins program?

b. Construct a 99% confidence interval estimate of the mean weightloss for all such subjects.

c. Does the Atkins program appear to be effective? Is itpractical?

Answer 1

a) For this case the best point estimate of the population mean weight loss of all overweight adults who follow the Atkins program is the sample mean:

`\hat \mu = \bar X = 2.1`

b)
`2.1-2.708(4.8)/(√(40))=0.0448`

`2.1+2.708(4.8)/(√(40))=4.1552`

c) Since the confidence interval contains only positive values we can conclude that the program is effective and is enough evidence at 1% of significance to conclude that the true weigth loss is higher than 0.

Explanation:

Data given

`\bar X=2.1` represent the sample mean for the weigth loss

`\mu` population mean

s=4.8 represent the sample standard deviation

n=40 represent the sample size

Part a

For this case the best point estimate of the population mean weight loss of all overweight adults who follow the Atkins program is the sample mean:

`\hat \mu = \bar X = 2.1`

Part b

The confidence interval for the true mean is given by the following formula:

`\bar X \pm t_(\alpha/2)(s)/(√(n))` (1)

The degrees of freedom are given by:

`df=n-1=40-1=39`

The Confidence level is 0.99 or 99%, the significance is
`\alpha=0.01` and
`\alpha/2 =0.005`, the critical value for this case is
`t_(\alpha/2)=2.708`

The confidence interval is given by:

`2.1-2.708(4.8)/(√(40))=0.0448`

`2.1+2.708(4.8)/(√(40))=4.1552`

Part c

Since the confidence interval contains only positive values we can conclude that the program is effective and is enough evidence at 1% of significance to conclude that the true weigth loss is higher than 0.