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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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” (Source: www.pollingreport.com/ life.htm).

a. Calculate the standard error of the sample proportion.

b. Calculate a 95% confidence interval that estimates the proportion of U.S. families with children who always have the television on when they eat dinner as a family.

c. Calculate a 90% confidence interval for this situation.

d. Write a sentence that interprets the confidence interval found in part (c).

Answered Same DayDec 25, 2021

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

124 Votes

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95% Confidence Interval for p
p̂ = 0.33,n = 254
Significance 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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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”...

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

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