# In a survey of 2,255 randomly selected US adults (age 18 or older), 1,787 of them use the...

## Question:

In a survey of 2,255 randomly selected US adults (age 18 or older), 1,787 of them use the Internet regularly. Of the Internet users, 1,054 use a social networking site.

Find a 90% confidence interval for each of the following proportions:

(a) Proportion of US adults who use the Internet regularly. Round your answers to three decimal places.

(b) Proportion of US adult Internet users who use a social networking site. Round your answers to three decimal places.

(c) Proportion of all US adults who use a social networking site. Round your answers to three decimal places.

(d) Use the confidence interval to estimate whether it is plausible to estimate that 50% of all US adults now use a social networking site.

## Confidence interval of proportions

The technique of confidence interval estimation used to determine the estimated range in which the future value of a parameter of the population may lie. The confidence interval gives two bounds, namely the lower and upper limits of the confidence interval. This interval width gets longer with the decrease in the level of confidence.

Given Information

{eq}{x_1} {/eq}: Proportion of US adults who use the Internet regularly.

{eq}\;{n_1} {/eq}: Population of surveyed US adults.

The samples are obtained independently using simple random sampling.

{eq}\begin{align*} {x_1} &= 1787\\ {n_1} &= 2255 \end{align*} {/eq}

The tabulated value of Z at 10% level of significance is calculated using standard normal table and is given by:

{eq}{Z_{0.10}} = 1.645 {/eq}

The sample proportion for first sample that is the proportion of US adults who use the Internet regularly is given by:

{eq}\begin{align*} {{\hat p}_1} &= \dfrac{{{x_1}}}{{{n_1}}}\\ &= \dfrac{{1787}}{{2255}}\\ &= 0.7924611973 \end{align*} {/eq}

a) The 90% confidence interval for Proportion of US adults who use the Internet regularly is given by:

{eq}\begin{align*} CI &= \left( {{{\hat p}_1} \pm {Z_{0.10}}\sqrt {\dfrac{{{{\hat p}_1}\left( {1 - {{\hat p}_1}} \right)}}{{{n_1}}}} } \right)\\ &= \left( {0.7924611973 \pm 1.645\sqrt {\dfrac{{0.7924611973 \times \left( {1 - 0.7924611973} \right)}}{{2255}}} } \right)\\ &= \left( {0.77841265,0.80650974} \right) \end{align*} {/eq}

The 90% confidence interval for Proportion of US adults who use the Internet regularly is (0.77841265,0.80650974).

{eq}{x_2} {/eq}: Proportion of US adult Internet users who use a social networking site.

{eq}{n_2} {/eq}: Population of Internet users in US.

The sample proportion for second sample that is the proportion of US adult Internet users who use a social networking site is given by:

{eq}\begin{align*} {{\hat p}_2} &= \dfrac{{{x_2}}}{{{n_2}}}\\ &= \dfrac{{1054}}{{1787}}\\ &= 0.589815333 \end{align*} {/eq}

b) The 90% confidence interval for proportion of US adult Internet users who use a social networking site is given by:

{eq}\begin{align*} CI &= \left( {{{\hat p}_2} \pm {Z_{0.10}}\sqrt {\dfrac{{{{\hat p}_2}\left( {1 - {{\hat p}_2}} \right)}}{{{n_2}}}} } \right)\\ &= \left( {0.589815333 \pm 1.645\sqrt {\dfrac{{0.589815333 \times \left( {1 - 0.589815333} \right)}}{{1787}}} } \right)\\ &= \left( {0.5706749,0.6089557} \right) \end{align*} {/eq}

The 90% confidence interval for proportion of US adult Internet users who use a social networking site is \left( {0.5706749,0.6089557} \right)

{eq}{x_3} {/eq}: Proportion of US adult who use a social networking site.

{eq}{n_3} {/eq}: Population of surveyed US adults.

The sample proportion for third sample that is the Proportion of all US adults who use a social networking site is given by:

{eq}\begin{align*} {{\hat p}_3} &= \dfrac{{{x_3}}}{{{n_3}}}\\ &= \dfrac{{1054}}{{2255}}\\ &= 0.4674057 \end{align*} {/eq}

c) The 90% confidence interval for Proportion of all US adults who use a social networking site is given by:

{eq}\begin{align*} CI &= \left( {{{\hat p}_3} \pm {Z_{0.10}}\sqrt {\dfrac{{{{\hat p}_3}\left( {1 - {{\hat p}_3}} \right)}}{{{n_3}}}} } \right)\\ &= \left( {0.4674057 \pm 1.645\sqrt {\dfrac{{0.4674057 \times \left( {1 - 0.4674057} \right)}}{{2255}}} } \right)\\ &= \left( {0.4501220,0.4846894} \right) \end{align*} {/eq}

The 90% confidence interval for Proportion of all US adults who use a social networking site is (0.4501220, 0.4846894).

d) The 90% confidence interval for Proportion of all US adults who use a social networking site is (0.4501220, 0.4846894).

Using the confidence interval (0.4501220, 0.4846894), we estimate that it is not plausible to estimate that 50% of all US adults now use a social networking site because 0.50 does not lie in the interval (0.4501220, 0.4846894) as both upper and lower limits are less than 0.50.

Finding Confidence Intervals with the Normal Distribution

from Statistics 101: Principles of Statistics

Chapter 9 / Lesson 3
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