Respuesta :
Confidence interval = p+/- standard error
Where;
p = percentage of he population using internet = 57% = 0.57
Standard error = Z*Sqrt [p(1-p)/N)]; where N = sample size = 1025, Z at 99% confidence = 2.58
Substituting;
Standard error = 2.58*Sqrt [0.57(1-0.57)/1025] ≈ 0.04
Then,
Confidence interval for the true population using the internet = 0.57 +/- 0.04 = [(0.57-0.04),(0.57+0.04)] = [0.53,0.61] = [53%, 61%].
Where;
p = percentage of he population using internet = 57% = 0.57
Standard error = Z*Sqrt [p(1-p)/N)]; where N = sample size = 1025, Z at 99% confidence = 2.58
Substituting;
Standard error = 2.58*Sqrt [0.57(1-0.57)/1025] ≈ 0.04
Then,
Confidence interval for the true population using the internet = 0.57 +/- 0.04 = [(0.57-0.04),(0.57+0.04)] = [0.53,0.61] = [53%, 61%].
The [tex]99\%[/tex] confidence interval for the true proportion of adults who use the Internet for shopping is \left( {0.53,0.61} \right).
Further Explanation:
Proportion of true mean follows normal distribution with mean [tex]np[/tex] and variance [tex]np(1-p)[/tex].
[tex]\boxed{p \sim N\left( {np,np\left( {1 - p} \right)} \right)}[/tex]
Confidence interval of proportion of mean can be expressed as,
[tex]{\text{Confidence interval}} = \left( {p - {\text{ME}},p + {\text{ME}}} \right)[/tex]
Given:
The total number of adults selected is [tex]1025[/tex].
[tex]57\%[/tex] of adults use internet for shopping from [tex]1025[/tex] adults.
[tex]p = 0.57[/tex]
Calculation:
The level of significance is [tex]1\%[/tex].
The value of [tex]{Z_{0.5\% }}[/tex] is [tex]2.58[/tex] that can be obtained from the standard normal table.
The formula for margin of error is,
[tex]\boxed{{\text{ME}} = {Z_{\frac{\alpha }{2}}} \times \sqrt {\frac{{p\left( {1 - p} \right)}}{n}} }[/tex]
The margin of error can be obtained as,
[tex]\begin{aligned} {\text{ME}} &= 2.58 \times \sqrt {\frac{{0.57\left( {1 - 0.57} \right)}}{{1025}}} \\ &= 2.58 \times \sqrt {\frac{{0.57 \times 0.43}}{{1025}}} \\ &= 0.04 \\ \end{aligned}[/tex]
The confidence interval can be obtained as,
[tex]\begin{aligned} {\text{Confidence interval}} &= \left( {0.57 - 0.04,0.57 + 0.04} \right) \\ &= \left( {0.53,0.61} \right) \\ \end{aligned}[/tex]
The [tex]99\%[/tex] confidence interval for the true proportion of adults who use the Internet for shopping is [tex]\left( {0.53,0.61} \right)[/tex].
Learn more:
1. Learn more about normal distribution https://brainly.com/question/12698949
2. Learn more about standard normal distribution https://brainly.com/question/13006989
3. Learn more about confidence interval of mean https://brainly.com/question/12986589
Answer details:
Grade: College
Subject: Statistics
Chapter: Confidence Interval
Keywords: Z-score, Z-value, binomial distribution, standard normal distribution, standard deviation, criminologist, test, measure, probability, low score, mean, repeating, indicated, normal distribution, percentile, percentage, undesirable behavior, proportion, empirical rule.
