1. In a recent study, the researchers want to find the information about the percentage ofadults who get their news online.a. In a Harris poll of 3026 adults, 38% said that they prefer to get their news online.Construct a 95% confidence interval estimate of the percentage of all adults whosay that they prefer to get their news online. Can we safely say that fewer than50% of adults prefer to get their news online?b. If we want to conduct another survey, how many adults must we survey in orderto be 90% confident that we are within four percentage points of the populationpercentage? Assume that we have no prior information about the populationproportion.

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Asked Oct 17, 2019
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1. In a recent study, the researchers want to find the information about the percentage of
adults who get their news online.
a. In a Harris poll of 3026 adults, 38% said that they prefer to get their news online.
Construct a 95% confidence interval estimate of the percentage of all adults who
say that they prefer to get their news online. Can we safely say that fewer than
50% of adults prefer to get their news online?
b. If we want to conduct another survey, how many adults must we survey in order
to be 90% confident that we are within four percentage points of the population
percentage? Assume that we have no prior information about the population
proportion.
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1. In a recent study, the researchers want to find the information about the percentage of adults who get their news online. a. In a Harris poll of 3026 adults, 38% said that they prefer to get their news online. Construct a 95% confidence interval estimate of the percentage of all adults who say that they prefer to get their news online. Can we safely say that fewer than 50% of adults prefer to get their news online? b. If we want to conduct another survey, how many adults must we survey in order to be 90% confident that we are within four percentage points of the population percentage? Assume that we have no prior information about the population proportion.

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Expert Answer

Step 1

Given data

n = 3026

proportion of people who prefer to get their news online = 0.38

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Confidence level= 0.95 Significance level = a = 1 - 0.95 = 0.05 Zo.05 Z0.025= -1.96 (from excel = NORM. S. INV(0.025)) Margin of error is given by Р(1-р) Error = Za п 0.38(1 0.38) Error = 1.96 = 0.0173 3026 Confidence interval is given by Р(1 - р) p+Za 0.38(1 0.38) = 0.38 t1.96 (0.3627,0.3973) 3026

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Step 2

b)
population proportion is not given,
so, we assume equal proportion i.e. p =...

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Significance level = a = 1 - 0.90 = 0.1 Zo.1 Zo.05 1.645 2 (Using Excel NORM.S.INV (0.05)) Margin of error formula is given by p(1-p) Za X E = Simplifying the above formula 2 1.645 2 п %3D p(1 — р) х E 0.5 х 0.5 х = 422.816 423 0.04

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