Question

select the correct answer.
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 statement holds true?
a. as the sample size is too small, the margin of error is 0.203.

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 appropriately large, the margin of error is 0.171.

Explanation:

Step1: Recall Margin of Error Formula

The formula for the margin of error (ME) for a proportion is \( ME = z_{\alpha/2} \times \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \), where \( z_{\alpha/2} \) is the z - score for the confidence level, \( \hat{p} \) is the sample proportion, and \( n \) is the sample size. For a 95% confidence level, \( z_{\alpha/2}=1.96 \) (from standard normal distribution tables).

Step2: Identify Values

We have \( \hat{p} = 0.65 \), \( n = 30 \), and \( z_{\alpha/2}=1.96 \).

Step3: Calculate the Margin of Error

First, calculate \( \hat{p}(1 - \hat{p})=0.65\times(1 - 0.65)=0.65\times0.35 = 0.2275 \).
Then, calculate \( \frac{\hat{p}(1 - \hat{p})}{n}=\frac{0.2275}{30}\approx0.007583 \).
Next, take the square root: \( \sqrt{\frac{0.2275}{30}}\approx\sqrt{0.007583}\approx0.0871 \).
Finally, multiply by \( z_{\alpha/2} \): \( ME = 1.96\times0.0871\approx0.1707\approx0.171 \). But wait, the sample size \( n = 30 \) is relatively small (a common rule of thumb is that \( np\geq5 \) and \( n(1 - p)\geq5 \), here \( np = 30\times0.65 = 19.5\geq5 \) and \( n(1 - p)=30\times0.35 = 10.5\geq5 \), but the sample size of 30 is small compared to the population size of 450? Wait, no, the formula for margin of error for proportions when the population is large (or when sampling with replacement) is as above. However, when the sample size is small (compared to the population, usually if \( n/N>0.05 \), we use the finite population correction factor, but \( n = 30 \), \( N = 450 \), \( n/N=\frac{30}{450}\approx0.0667>0.05 \), so we should use the finite population correction factor \( \sqrt{\frac{N - n}{N - 1}} \).
Let's recalculate with the finite population correction factor. The finite population correction factor \( FPC=\sqrt{\frac{450 - 30}{450 - 1}}=\sqrt{\frac{420}{449}}\approx\sqrt{0.9354}\approx0.9672 \).
Now, the standard error \( SE=\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}\times FPC\approx0.0871\times0.9672\approx0.0842 \).
Then, \( ME = z_{\alpha/2}\times SE=1.96\times0.0842\approx0.165 \). But the options given have 0.203, 0.171. Wait, maybe we made a mistake in the initial approach. Wait, let's check the options again. Wait, maybe the question is considering the sample size as too small. Wait, the sample size \( n = 30 \) is small. Let's recalculate without the finite population correction factor:
\( ME = 1.96\times\sqrt{\frac{0.65\times0.35}{30}}=1.96\times\sqrt{\frac{0.2275}{30}}=1.96\times\sqrt{0.007583}\approx1.96\times0.0871\approx0.171 \). But the sample size of 30 is small (since a larger sample size gives a more accurate estimate). Wait, the options:
Option A: Says sample size too small, ME = 0.203. Let's recalculate ME with maybe a wrong z - score? Wait, if we use \( z = 2.576 \) (for 99% confidence), but no, it's 95% confidence. Wait, maybe the question has a different approach. Wait, let's check the calculation again.
Wait, \( \hat{p}=0.65 \), \( n = 30 \)
\( \sqrt{\frac{0.65\times0.35}{30}}=\sqrt{\frac{0.2275}{30}}\approx\sqrt{0.007583}\approx0.0871 \)
\( ME = 1.96\times0.0871\approx0.171 \). But the sample size of 30 is small. Wait, the options:
Option C says "As the sample size is too small, the margin of error cannot be trusted." But let's check the calculation of ME again. Wait, maybe the question has a typo or we misread. Wait, if we calculate \( ME \) with \( z = 2.33 \) (for 98% confidence) or other? No, 95% confidence is 1.96. Wait, maybe the sample size is considered too small because 30 is small. But let's check the options:
Option D: "As the sample size is appropriately large, the margin of error is 0.171." But 30 is not a large sample size (usually \( n\geq30 \) is a rule of thumb for the Central Limit Theorem to apply approximately, but 30 is the lower end). Wait, the Central Limit Theorem for proportions: the sampling distribution of the sample proportion is approximately normal if \( np\geq5 \) and \( n(1 - p)\geq5 \), which is satisfied here. But the sample size of 30 is small. However, when we calculated, the ME is approximately 0.171. But let's check the calculation again. Wait, maybe the formula was applied incorrectly. Wait, another way: maybe the question uses a different z - score or has a miscalculation. Wait, if we use \( z = 2.0 \) (approximate for 95% confidence), then \( ME = 2\times\sqrt{\frac{0.65\times0.35}{30}}=2\times0.0871 = 0.1742\approx0.171 \). But the sample size of 30: is it appropriately large? A sample size of 30 is the minimum for the Central Limit Theorem to start to apply, but it's not a very large sample. However, among the options, option D says "As the sample size is appropriately large, the margin of error is 0.171". Wait, but our calculation gave ME≈0.171. But let's check the other options:
Option A: ME = 0.203. Let's calculate ME with \( z = 2.33 \) (98% confidence) or something else. Wait, \( ME = 2.33\times\sqrt{\frac{0.65\times0.35}{30}}=2.33\times0.0871\approx0.203 \). But for 95% confidence, z = 1.96. So if the sample size is too small, maybe we can't trust the z - score? Wait, the key here is: the sample size \( n = 30 \). A common rule is that if \( n<30 \), it's a small sample, but here \( n = 30 \). However, the population size is 450, and \( n = 30 \) is 6.67% of the population. The formula for margin of error when the population is finite is \( ME=z_{\alpha/2}\times\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}\times\frac{N - n}{N - 1}} \). Let's recalculate with \( N = 450 \):
\( \frac{N - n}{N - 1}=\frac{450 - 30}{450 - 1}=\frac{420}{449}\approx0.9354 \)
\( \sqrt{\frac{0.65\times0.35}{30}\times0.9354}=\sqrt{\frac{0.2275\times0.9354}{30}}=\sqrt{\frac{0.2138}{30}}\approx\sqrt{0.007127}\approx0.0844 \)
\( ME = 1.96\times0.0844\approx0.1654\approx0.165 \). But the options have 0.171 and 0.203. Wait, maybe the question does not use the finite population correction factor. Then \( ME = 1.96\times\sqrt{\frac{0.65\times0.35}{30}}\approx1.96\times0.0871\approx0.171 \). Now, is the sample size appropriately large? The Central Limit Theorem says that as the sample size increases, the sampling distribution of the sample proportion approaches a normal distribution. A sample size of 30 is considered the minimum for the approximation to be reasonable (for the Central Limit Theorem). But compared to the population size of 450, 30 is small, but the key is the sample proportion's distribution. Since \( np = 19.5\geq5 \) and \( n(1 - p)=10.5\geq5 \), the sampling distribution is approximately normal. However, the sample size of 30 is not very large. But among the options, option D says "As the sample size is appropriately large, the margin of error is 0.171". Our calculation gives ME≈0.171. Option A says ME = 0.203, which would be if z = 2.33 (98% confidence), but it's 95% confidence. Option C says the margin of error cannot be trusted because the sample size is too small, but since \( np\) and \( n(1 - p)\) are greater than 5, the approximation is reasonable. Wait, maybe the question considers \( n = 30 \) as appropriately large (since \( n\geq30 \) is a rule of thumb for CLT). So the correct option is D.

Answer:

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