In a study of 1910 schoolchildren in Australia, 1050 children indicated that they normally watch TV...

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In a study of 1910 schoolchildren in Australia, 1050 childrenindicated that they normally watch TV before school in the morning.(Interestingly, only 35% of the parents said their children watchedTV before school!)
(a)
Construct a 95% confidence interval for the true proportion ofAustralian children who say they watch TV before school. (Roundyour answers to three decimal places.)
(_,____)
What assumption about the sample must be true for the method usedto construct the interval to be valid?(b)
The 1910 schoolchildren used in the study formed a random samplefrom the population of children in Australia who normally watch TVbefore school in the morning.
The 1050 children who indicated that they normally watch TVbefore school in the morning formed a random sample from thepopulation of schoolchildren inAustralia.
The 1910 schoolchildren used in the study formed a random samplefrom the population of schoolchildren in Australia.
The 1050 children who indicated that they normally watch TVbefore school in the morning formed a random sample from thepopulation of children in Australia who normally watch TV beforeschool in the morning.

Answer & Explanation Solved by verified expert

4.1 Ratings (445 Votes)

Since we are testing for school chidren who told they watched tvbefore school we will use the proportion of children who indicatedthey watch tv and not what their parents saidThis is confidence interval for true binomial See Answer

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