Question

A health promotion campaign in schools aimed to teach children to choose healthy foods when selecting snacks.

A random sample of 120 children who had heard the campaign were selected and asked what kind of snacks they had eaten on the previous day.

A separate random sample of 110 children from schools where the campaign has not been run were asked the same question.

Of the 120 children who had heard the campaign, 75 had chosen healthy snacks.

Of the 110 who did not hear the campaign, 47 chose healthy snacks.

Find an estimate for the true difference in the proportion of children choosing healthy snacks in the two groups (Campaign - No campaign).

and the confidence interval.

What I've tried is 75/120 -47/110 = 0.198. This is the difference of the mean, but how can I find the standard deviation to achieve the confidence interval?

Answer 1

i am denoting the Campaign sample by group 1 and No campaign sample by group 2.

the given data and necessary calculations be:-

the standard deviation needed to achieve the confidence interval be:-

now you can easily find out the confidence interval by using the formula:-

*** if you have any doubts about the problem please mention it in the comment box... if satisfied please LIKE.