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Nikhil asked 120 randomly chosenmoviegoers to watch a clip of an upcoming movie and choose the best title. The title "Everything" was chosen by 36 of the moviegoers as the preferred title. To the nearest percent, with a confidence level of 90% (z*-score 1.645), what is the confidence interval for the proportion of moviegoers who preferred the title "Everything"? E = z* and C = + E between 4% and 10% between 7% and 30% between 23% and 37% between 29% and 43%

Respuesta :

Answer:

C. between 23% and 37%

Explanation:

Given:

The total viewers(n) = 120

The total number that chooses the title  "Everything” = 36

Therefore,

The mean proportion is p = 36/120 = 0.3.

Let's find the standard deviation of the proportion

=√(p*(1-p)/n)

= √(0.3*0.7/120)

= 0.0418.

Next step is to multiply 0.0418 by the z-score of 1.645 to get the deviation

The confidence interval

= (0.3 - 0.0688, 0.3 + 0.0688)

=(0.2312, 0.3688).

=(23.12%, 36.88%)

myspamlevi

Answer:

C on edge

Explanation:

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