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.
Accepted Solution
A:
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 = 0.65So, 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