Question

In a Gallup poll, 1025 randomly selected adults were surveyed and 57% of them said that they used the Internet for shopping at least a few times a year. Construct a 99% confidence interval for the true proportion of adults who use the Internet for shopping.

Answer 1

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%].

Answer 2

The $99\%$ 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 $np$ and variance $np(1-p)$.

$\boxed{p \sim N\left( {np,np\left( {1 - p} \right)} \right)}$

Confidence interval of proportion of mean can be expressed as,

${\text{Confidence interval}} = \left( {p - {\text{ME}},p + {\text{ME}}} \right)$

Given:

The total number of adults selected is $1025$.

$57\%$ of adults use internet for shopping from $1025$ adults.

$p = 0.57$

Calculation:

The level of significance is $1\%$.

The value of ${Z_{0.5\% }}$ is $2.58$ that can be obtained from the standard normal table.

The formula for margin of error is,

$\boxed{{\text{ME}} = {Z_{\frac{\alpha }{2}}} \times \sqrt {\frac{{p\left( {1 - p} \right)}}{n}} }$

The margin of error can be obtained as,

$\begin{aligned} {\text{ME}} &= 2.58 \times \sqrt {\frac{{0.57\left( {1 - 0.57} \right)}}{{1025}}}  \\  &= 2.58 \times \sqrt {\frac{{0.57 \times 0.43}}{{1025}}}  \\   &= 0.04 \\ \end{aligned}$

The confidence interval can be obtained as,

$\begin{aligned}  {\text{Confidence interval}} &= \left( {0.57 - 0.04,0.57 + 0.04} \right) \\  &= \left( {0.53,0.61} \right) \\ \end{aligned}$

The $99\%$ confidence interval for the true proportion of adults who use the Internet for shopping is $\left( {0.53,0.61} \right)$.

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.