Respuesta :
Answer:
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:
[tex]\hat \mu = \bar X = 2.1[/tex]
b) [tex]2.1-2.708\frac{4.8}{\sqrt{40}}=0.0448[/tex]
[tex]2.1+2.708\frac{4.8}{\sqrt{40}}=4.1552[/tex]
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.
Step-by-step explanation:
Data given
[tex]\bar X=2.1[/tex] represent the sample mean for the weigth loss
[tex]\mu[/tex] 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:
[tex]\hat \mu = \bar X = 2.1[/tex]
Part b
The confidence interval for the true mean is given by the following formula:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
The degrees of freedom are given by:
[tex]df=n-1=40-1=39[/tex]
The Confidence level is 0.99 or 99%, the significance is [tex]\alpha=0.01[/tex] and [tex]\alpha/2 =0.005[/tex], the critical value for this case is [tex]t_{\alpha/2}=2.708[/tex]
The confidence interval is given by:
[tex]2.1-2.708\frac{4.8}{\sqrt{40}}=0.0448[/tex]
[tex]2.1+2.708\frac{4.8}{\sqrt{40}}=4.1552[/tex]
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.
