Respuesta :
Answer:
The 99% confidence interval for the population proportion is between (0.5341, 0.5843). The interpretation is that we are 99% sure that the true proportion of adults who have started paying bills online in the last year is between these two values.
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which
z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].
For this problem, we have that:
In a survey of 2591 adults, 1449 say they have started paying bills online in the last year. This means that [tex]n = 2591, p = \frac{1449}{2591} = 0.5592[/tex]
99% confidence level
So [tex]\alpha = 0.01[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.01}{2} = 0.995[/tex], so [tex]Z = 2.575[/tex].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5592 - 2.575\sqrt{\frac{0.5592*0.4408}{2591}} = 0.5341[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5592 + 2.575\sqrt{\frac{0.5592*0.4408}{2591}} = 0.5843[/tex]
The 99% confidence interval for the population proportion is between (0.5341, 0.5843). The interpretation is that we are 99% sure that the true proportion of adults who have started paying bills online in the last year is between these two values.
Answer:
The 99% confidence interval for the portion is (0.553, 0.603). It means that the 99% of the population portion is contained in that interval.
Step-by-step explanation:
1. Calculate the probability of any adult started paying bills online in the las year. In this case, [tex]N=2591[/tex] and [tex]n=1449[/tex]
[tex]p=\frac{n}{N}=\frac{1449}{2591}=0,59[/tex]
2. The significance level of 99% confident is:
[tex]99%=1-\alpha\\\alpha=1-0.99\\\alpha=0.1[/tex]
Then, according to the normal distribution chart, the z-value for [tex]\alpha/2=0.05[/tex] is 2.58
3. Estimate Standard Error of p:
[tex]SE=\sqrt{\frac{px(1-p)}{N} } \\SE=\sqrt{\frac{px(1-0.579)}{2591} } \\SE=0,0097[/tex]
4. Estimate Error E for [tex]Z_{\frac{\alpha}{2} }[/tex]
[tex]E=Z_{\alpha/2}x\sqrt{\frac{px(1-p)}{N} } \\E=Z_{\0.1/2}x\sqrt{\frac{0.579x(1-0.579)}{2591} } \\E=2.58x0.0097=0.25[/tex]
5. Construct the confidence interval:
CI=p ± E
CI=0.579 ± 0.025
6. Calculate Lower and Upper Limits
Lower Liimit= p - E
Lower Limit=0.579-0.025 =0.553
Upper Limit= p + E
Upper Limit= 0.579+ 0.025= 0.603
7. The 99% confidence interval for the portion is (0.553, 0.603). It means that the 99% of the population portion is contained in that interval.
