Respuesta :
Answer:
a) [tex] \hat p=\frac{914}{4579}=0.1996[/tex] and represent the proportion of American adults who earned money by selling osmething online the previous year.
[tex]0.1996 - 2.58 \sqrt{\frac{0.1996(1-0.1996)}{4579}}=0.184[/tex]
[tex]0.1996 + 2.58 \sqrt{\frac{0.1996(1-0.1996)}{4579}}=0.215[/tex]
And the 99% confidence interval would be given (0.184;0.215).
b) For this case we see that the value of 0.2 or 20% is included on the confidence interval at 99% of confidence so then we don't have enough evidence to conclude that the proportion is less than 0.2 since the upper limit for the confidence interval is higher than 0.2 and the lower limit is lower than 0.2.
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".
The population proportion have the following distribution
[tex]p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})[/tex]
Part a
The confidence interval would be given by this formula
[tex]\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
For the 99% confidence interval the value of [tex]\alpha=1-0.99=0.01[/tex] and [tex]\alpha/2=0.005[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.
[tex]z_{\alpha/2}=2.58[/tex]
The estimated propotion for this case is given by:
[tex] \hat p=\frac{914}{4579}=0.1996[/tex] and represent the proportion of American adults who earned money by selling osmething online the previous year.
And the sample size on this case is n =4579
And replacing into the confidence interval formula we got:
[tex]0.1996 - 2.58 \sqrt{\frac{0.1996(1-0.1996)}{4579}}=0.184[/tex]
[tex]0.1996 + 2.58 \sqrt{\frac{0.1996(1-0.1996)}{4579}}=0.215[/tex]
And the 99% confidence interval would be given (0.184;0.215).
Part b
For this case we see that the value of 0.2 or 20% is included on the confidence interval at 99% of confidence so then we don't have enough evidence to conclude that the proportion is less than 0.2 since the upper limit for the confidence interval is higher than 0.2 and the lower limit is lower than 0.2.
99% confidence interval is given by (0.184,0.215) and this can be determined by using the confidence interval formula.
Given :
In a random sample of 4579 American adults, 914 reported that they earned money by selling something online in the previous year.
A) The formula to find the confidence interval is:
[tex]\rm CI = \hat{p}\pm z_{\frac{\alpha }{2}} \sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex] ----- (1)
[tex]\alpha = 1 - 0.99 = 0.01[/tex]
[tex]\dfrac{\alpha }{2} = 0.005[/tex]
So, [tex]z_{\frac{\alpha }{2}} = 2.58[/tex]
Estimated proportion is given by:
[tex]\rm \hat{p}= \dfrac{914}{4579}=0.1996[/tex]
Now, put the values of all known terms in equation (1) in order to determine the confidence interval.
[tex]\rm CI = 0.1996 \pm 2.58 \sqrt{\dfrac{0.1996(1-0.1996)}{4579}}[/tex]
Now, simplify the above equation.
Cl = 0.184 , 0.215
B) The value of 0.2 or 20% is included on the confidence interval at 99% of confidence so then we don't have enough evidence to conclude that less than 20% earned money by selling something online.
For more information, refer to the link given below:
https://brainly.com/question/2561151
