Eric randomly surveyed 150 adults from a certain city and asked which team in a contest they were rooting for, either north high school or south high school. from the results of his survey, eric obtained a 95% confidence interval of (0.52,0.68) for the proportion of all adults in the city rooting for north high. what proportion of the 150 adults in the survey said they were rooting for north high school?

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Maria Page

Business

26 June, 21:43

Eric randomly surveyed 150 adults from a certain city and asked which team in a contest they were rooting for, either north high school or south high school. from the results of his survey, eric obtained a 95% confidence interval of (0.52,0.68) for the proportion of all adults in the city rooting for north high. what proportion of the 150 adults in the survey said they were rooting for north high school?

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Grey · Answer 1

Interval at 95% confidence = p + / - Z sqrt (p (1-p) / n)

Upper interval = > 0.68=p+1.96 sqrt (p (1-p) / n)

Lower interval = > 0.52=p-1.96 sqrt (p (1-p) / n)

Putting sqrt term to be "q"

0.68=p+1.96q

0.52=p-1.96q

Adding the two equations to solve for p (proportion of 150 adults rootng for North high school);

1.2 = 2p = > p=0.6