Question

17
select the correct answer.
the quantity of soda consumed per week at a soda fountain is 34 liters, with a standard deviation of 5.8 liters. if any 80 weeks were sampled randomly, what would be the margin of error of the sample mean?
a. 1.944 liters
b. 1.297 liters
c. 0.994 liters
d. 0.648 liters
reset next

Explanation:

Step1: Recall Margin of Error Formula

The formula for the margin of error (ME) for the sample mean when the population standard deviation ($\sigma$) is known is $ME = z^* \times \frac{\sigma}{\sqrt{n}}$. For a 95% confidence interval (common, and we assume this here as it's not specified otherwise), $z^* \approx 1.96$. Here, $\sigma = 5.8$ liters, $n = 80$.

Step2: Calculate Standard Error

First, calculate the standard error (SE) which is $\frac{\sigma}{\sqrt{n}}$. So, $\sqrt{80} \approx 8.944$. Then, $\frac{5.8}{8.944} \approx 0.6484$.

Step3: Calculate Margin of Error

Multiply the standard error by $z^*$ (1.96). So, $ME = 1.96 \times 0.6484 \approx 1.291$. Wait, maybe we use $z = 1.96$? Wait, let's recalculate. Wait, $\sqrt{80} = 4\sqrt{5} \approx 8.94427$. Then $\frac{5.8}{8.94427} \approx 0.6484$. Then $1.96 \times 0.6484 \approx 1.291$, which is close to option B (1.297, maybe due to using a more precise $z$ or rounding). Wait, maybe the problem uses $z = 1.96$ or maybe a different approach? Wait, let's check again. Wait, maybe the question is using the formula for margin of error with $z = 1.96$ (95% CI). So:

$ME = 1.96 \times \frac{5.8}{\sqrt{80}}$

Calculate $\sqrt{80} \approx 8.9443$

$\frac{5.8}{8.9443} \approx 0.6484$

$1.96 \times 0.6484 \approx 1.291$, which is approximately 1.297 (maybe due to more precise $z$ or rounding during steps). So the closest is option B.

Answer:

B. 1.297 liters