Question

Due to the popularity of instant messaging and social​ networking, informal elements such as emoticons​ (e.g., the symbol​ ":)" to represent a​ smile) and abbreviations​ (e.g., "LOL" for​ "laughing out​ loud") have worked their way into​ teenagers' school writing assignments. A survey interviewed 900 randomly selected teenagers by telephone on their writing habits.​ Overall, 442 of the teenagers admitted using at least one informal element in school writing assignments. Based on the survey​ results, construct a 90​% confidence interval for the proportion of all teenagers who have used at least one informal element in school writing assignments. Give a practical interpretation of the interval.

Answer 1

`0.491 - 1.64\sqrt{(0.491(1-0.491))/(900)}=0.464`

`0.491 + 1.64\sqrt{(0.491(1-0.491))/(900)}=0.518`

We are 90% confident that the true proportion of teenagers who have used at least one informal element in school writing assignments is between 0.464 and 0.518

Explanation:

We can begin finding the estimator for the proportion of interest:

`\hat p =(442)/(900)= 0.491`

The confidence level is 90% , and the significance level would be given by
`\alpha=1-0.90=0.1` and
`\alpha/2 =0.05`. And the critical value would be given by:

`z_(\alpha/2)=-1.64, z_(1-\alpha/2)=1.64`

The confidence interval for the mean is given by the following formula:

`\hat p \pm z_(\alpha/2)\sqrt{(\hat p (1-\hat p))/(n)}`

Replacing the info given we got:

`0.491 - 1.64\sqrt{(0.491(1-0.491))/(900)}=0.464`

`0.491 + 1.64\sqrt{(0.491(1-0.491))/(900)}=0.518`

We are 90% confident that the true proportion of teenagers who have used at least one informal element in school writing assignments is between 0.464 and 0.518