A coffee house owner knows that customers pour different amounts of coffee into their cups. She samples cups from 10 customers she believes to be representative of the customers and weighs the cups, finding a mean of 12.5 ounces and standard deviation of 0.5 ounces. Assuming these cups of coffee can be considered a random sample of all cups of coffee which of the following formulas gives a 95% confidence interval for the mean weight of all cups of coffee?
A) $ 12.5 \pm 2.228\left(\frac{0.5}{\sqrt{9}}\right) $

B) $ 12.5 \pm 2.262\left(\frac{0.5}{\sqrt{9}}\right) $

C) $ 12.5 \pm 2.228\left(\frac{0.5}{\sqrt{10}}\right) $

D) $ 12.5 \pm 2.262\left(\frac{0.5}{\sqrt{10}}\right) $

E) $ 12.5 \pm 1.96\left(\frac{0.5}{\sqrt{10}}\right) $

Question

A coffee house owner knows that customers pour different amounts of coffee into their cups. She samples cups from 10 customers she believes to be representative of the customers and weighs the cups, finding a mean of 12.5 ounces and standard deviation of 0.5 ounces. Assuming these cups of coffee can be considered a random sample of all cups of coffee which of the following formulas gives a 95% confidence interval for the mean weight of all cups of coffee? 
A) $ 12.5 \pm 2.228\left(\frac{0.5}{\sqrt{9}}\right) $

B) $ 12.5 \pm 2.262\left(\frac{0.5}{\sqrt{9}}\right) $

C) $ 12.5 \pm 2.228\left(\frac{0.5}{\sqrt{10}}\right) $

D) $ 12.5 \pm 2.262\left(\frac{0.5}{\sqrt{10}}\right) $

E) $ 12.5 \pm 1.96\left(\frac{0.5}{\sqrt{10}}\right) $

Quizplus · Accepted Answer

To find a confidence interval for the mean weight of all cups of coffee, we use the formula:
$$	ext{CI} = \bar{x} \pm t_{\alpha/2, n-1} \left(\frac{s}{\sqrt{n}}ight)$$
where $\bar{x}$ is the sample mean, $s$ is the sample standard deviation, $n$ is the sample size, and $t_{\alpha/2, n-1}$ is the t-score with $\alpha/2$ area to the right (and hence $\alpha/2$ area to the left).

We are given that the sample mean is $\bar{x} = 12.5$ ounces and the sample standard deviation is $s = 0.5$ ounces. We are also told that we want a 95% confidence interval, so our $\alpha$ is 0.05 and our $\alpha/2$ is 0.025. The sample size is $n = 10$.

To find $t_{\alpha/2, n-1}$, we look up the t-distribution with 9 degrees of freedom (since $n-1 = 9$) and a probability of 0.025 in each tail (since we want a two-tailed test). Using a t-table or calculator, we find that $t_{0.025, 9} = 2.262$.

Plugging in the values, we get:
$$	ext{CI} = 12.5 \pm 2.262\left(\frac{0.5}{\sqrt{10}}ight)$$
which simplifies to:
$$	ext{CI} = (12.26, 12.74)$$

Therefore, the best choice is D, with the correct formula for the 95% confidence interval.

A Coffee House Owner Knows That Customers Pour Different Amounts 12.5±2.228(0.59) 12.5 \pm 2.228\left(\frac{0.5}{\sqrt{9}}\right) 12.5±2.228(9​0.5​)

A Coffee House Owner Knows That Customers Pour Different Amounts $12.5 \pm 2.228\left(\frac{0.5}{\sqrt{9}}\right)$