Respuesta :
Answer:
The 99% lower confidence bound for the true average weight loss is 3.98 pounds.
This means that we can be 99% sure that the mean weight loss for all obese adults on a low-carb diet is positive.
Step-by-step explanation:
We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:
[tex]\alpha = \frac{1-0.99}{2} = 0.005[/tex]
Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].
So it is z with a pvalue of [tex]1-0.005 = 0.995[/tex], so [tex]z = 2.575[/tex]
Now, find the margin of error M as such
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.
[tex]M = 2.575*\frac{19}{\sqrt{66}} = 6.02[/tex]
The lower end of the interval is the sample mean subtracted by M. So it is 11 - 6.02 = 3.98 pounds
The 99% lower confidence bound for the true average weight loss is 3.98 pounds.
This means that we can be 99% sure that the mean weight loss for all obese adults on a low-carb diet is positive.
Answer:
[tex]11-2.385\frac{19}{\sqrt{66}}=5.422[/tex]
So the one side upper confidence interval is (5.422, infty)
And we can conclude that the minimum value that we expected for the true mean of the weight loss at 99% of confidence is 5.422
Step-by-step explanation:
Previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
[tex]\bar X=11[/tex] represent the sample mean
[tex]\mu[/tex] population mean (variable of interest)
[tex]s=19[/tex] represent the population standard deviation
n=66 represent the sample size
Solution to the problem
The degrees of freedom for this case are:
[tex] df = n-1= 66-1= 65[/tex]
Since the confidence is 0.99 or 99%, the value of [tex]\alpha=1-0.99=0.01[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.01,65)".And we see that [tex]t_{\alpha/2}=2.385[/tex]
Now we have everything in order to replace into formula (1):
[tex]11-2.385\frac{19}{\sqrt{66}}=5.422[/tex]
So the one side upper confidence interval is (5.422, infty)
And we can conclude that the minimum value that we expected for the true mean of the weight loss at 99% of confidence is 5.422
