Question

Based on a poll conducted by the Centers for Disease Control, 862 of 1013 randomly selected adults said that they always wear seat belts. Construct and interpret a 95% confidence interval for the proportion of adults who always wear seat belts.

Answer 1

The confidence interval is  0.028, 0.872

Given

862 of 1013 randomly selected adults  wear seat belts.

The 95% confidence interval for the proportion of adults who always wear seat belts.

What is confidence interval?

A Confidence Interval is defined as a range of values we are sure our true value lies within it.

What is 95% confidence interval?

A 95% confidence interval is defined as that if we take 100 different samples and compute a 95% confidence interval for each sample, then approximately 95 of the 100 confidence intervals will contain the true mean value.

= 862/1013 = 0.85

At a 95% confidence interval ,the critical value is $z_{0.025} = 1.96$

The 95% confidence interval for proportion is

$\widehat p +/- z0.025 * sqrt(\widehat p * (1 - \widehat p)/n)$

$= 0.85 +/- 1.96 * sqrt(0.85 * (1 - 0.85)/1013)$

$= 0.85 +/- 0.022$

= 0.028, 0.872

The sample should use at least 10 successes and 10 failures.

$n * \widehat p = 1013 * 0.85 = 861 > 10$

$n * (1 - \widehat p) = 1013 * 0.15 = 152 > 10$

As the sample size is larger, we can use an approximate formula for the standard error.

Learn more on Confidence intervals here:

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