Question

Joan conducted a study to see how common binge drinking is on her college campus. She defined "frequent binge drinking" as having five or more drinks in a row three or more times in the past two weeks. Out of 593 students who replied to her survey, 64 fit this criterion. Joan wants to construct a significance test for her data. She finds that the proportion of binge drinkers nationally is 13.1%. The z statistic for this data is __________.

Answer 1

Answer:

z = -1.66

Step-by-step explanation:

Z-statistic:

$z = \frac{X - p}{s}$

In which X is the found proportion.

p is the mean proportion.

$s = \sqrt{\frac{p(1-p)}{n}}$ is the standard error for the data.

Out of 593 students who replied to her survey, 64 fit this criterion.

This means that $X = \frac{64}{593} = 0.108$

She finds that the proportion of binge drinkers nationally is 13.1%.

This means that $p = 0.131$

Also

$s = \sqrt{\frac{0.131*0.869}{593}} = 0.014$

The z statistic for this data is

$z = \frac{X - p}{s}$

$z = \frac{0.108 - 0.131}{0.014}$

$z = -1.66$