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A restaurant owner wants to determine the effectiveness of his servers. The owner conducted a survey by asking 30 customers about the servers' effectiveness, on a day when the restaurant had 450 customers. The owner found that 65% of those surveyed were satisfied with the service offered. Assuming a 95% confidence level, which of the following statements holds true?
A. As the sample size is appropriately large, the margin of error is ±0.171.

B.As the sample size is appropriately large, the margin of error is ±0.203.

C.As the sample size is too small, the margin of error cannot be trusted.

D.As the sample size is too small, the margin of error is ±0.203.

Respuesta :

Answer:

option (C)

Step-by-step explanation:

Margin of error is determined by the formula :

[tex]\frac{z\cdot \sqrt{p\cdot (1-p)}}{\sqrt{(N-1)\cdot \frac{n}{N-n} }}[/tex]

z for 95% confidence level is 1.96 , N = 450 , n = 30 , p = 0.65

So, margin of error =

[tex]\frac{1.96\cdot \sqrt{0.65\cdot (1-0.65)}}{\sqrt{(450-1)\cdot \frac{30}{450-30} }}\\\frac{=1.96\cdot 0.477}{5.663}\\=0.16508\\=0.16508\cdot 100[/tex]

= 16.508 %

Which is very large because of the small sample size and therefore, the margin of error can't be trusted



mythforev16

Answer:

As the sample size is appropriately large, the margin of error is 0.171

Step-by-step explanation:

Got it right

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