In a Gallup poll done in 2007, a randomly selected sample of n = 254 U.S. parents was asked how often the television was on when they ate dinner as a family. About 33% of the sample said “always”...

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Answer To: In a Gallup poll done in 2007, a randomly selected sample of n = 254 U.S. parents was asked how...

Robert answered on Dec 25 2021
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95% Confidence Interval for p
p̂ = 0.33,n = 254
Signif
icance level = α = 1− confidence = 1 − 0.95 = 0.05
Critical z-value = zα/2 = z0.05/2 = z0.025 = 1.96 (From z table)
Standard error of p̂ : SE =

p̂× (1 − p̂)
n
=

0.33 × 0.67
254 ≈ 0.029504
Standard error = 0.0295
E = zα/2 ×

p̂× (1 − p̂)
n
= 1.96 × 0.029504 ≈ 0.057827
95% Confidence Interval is given by 0.33 ± 0.057827 :
Lower limit = p̂− E = 0.33 − 0.057827 ≈ 0.272173
Upper limit = p̂+ E = 0.33 + 0.057827 ≈...
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