The manager of a self-serve frozen yogurt shop, Purple Pear, knows that customers swirl different amounts of the creamy concoction into their cups. She samples cups from 10 costumers she believes to be representative of the customers and weighs the cups, finding a mean of 4.5 ounces and standard deviation of 0.25 ounces. Assuming these cups of frozen yogurt can be considered a random sample of all cups of frozen yogurt which of the following formulas gives a 95%
Confidence interval for the mean weight of all cups of frozen yogurt?
A) $ 4.5 \pm 2.228\left(\frac{0.25}{\sqrt{10}}\right) $

B) $ 4.5 \pm 2.228\left(\frac{0.25}{\sqrt{9}}\right) $

C) $ 4.5 \pm 2.262\left(\frac{0.25}{\sqrt{9}}\right) $

D) $ 4.5 \pm 2.262\left(\frac{0.25}{\sqrt{10}}\right) $

E) $ 4.5 \pm 1.969\left(\frac{0.25}{\sqrt{10}}\right) $

Question

The manager of a self-serve frozen yogurt shop, Purple Pear, knows that customers swirl different amounts of the creamy concoction into their cups. She samples cups from 10 costumers she believes to be representative of the customers and weighs the cups, finding a mean of 4.5 ounces and standard deviation of 0.25 ounces. Assuming these cups of frozen yogurt can be considered a random sample of all cups of frozen yogurt which of the following formulas gives a 95%
Confidence interval for the mean weight of all cups of frozen yogurt? 
A) $ 4.5 \pm 2.228\left(\frac{0.25}{\sqrt{10}}\right) $

B) $ 4.5 \pm 2.228\left(\frac{0.25}{\sqrt{9}}\right) $

C) $ 4.5 \pm 2.262\left(\frac{0.25}{\sqrt{9}}\right) $

D) $ 4.5 \pm 2.262\left(\frac{0.25}{\sqrt{10}}\right) $

E) $ 4.5 \pm 1.969\left(\frac{0.25}{\sqrt{10}}\right) $

Quizplus · Accepted Answer

We use the formula for a confidence interval for the mean of a population when the sample standard deviation is known:
$$\bar{x} \pm z_{\alpha/2} \left(\frac{\sigma}{\sqrt{n}}\right)$$
where $\bar{x}$ is the sample mean, $\sigma$ is the population standard deviation, $n$ is the sample size, and $z_{\alpha/2}$ is the z-score that corresponds to the desired level of confidence (in this case, 95%). Since the population standard deviation is not known, we use the sample standard deviation as an estimate, which gives us the formula:
$$\bar{x} \pm t_{\alpha/2, n-1} \left(\frac{s}{\sqrt{n}}\right)$$
where $s$ is the sample standard deviation.
We are given $\bar{x} = 4.5$, $s = 0.25$, and $n=10$. To find the appropriate t-value for a 95% confidence interval with 9 degrees of freedom (since we are estimating the population standard deviation from the sample standard deviation), we use a t-table or calculator and find $t_{0.025, 9} = 2.262$. Plugging in the numbers into the formula, we get:
$$4.5 \pm 2.262\left(\frac{0.25}{\sqrt{10}}\right)$$
which simplifies to:
$$4.5 \pm 0.204$$
Therefore, the 95% confidence interval for the mean weight of all cups of frozen yogurt is approximately $(4.296, 4.704)$. The correct answer is D.

The Manager of a Self-Serve Frozen Yogurt Shop, Purple Pear 4.5±2.228(0.2510) 4.5 \pm 2.228\left(\frac{0.25}{\sqrt{10}}\right) 4.5±2.228(10​0.25​)

The Manager of a Self-Serve Frozen Yogurt Shop, Purple Pear $4.5 \pm 2.228\left(\frac{0.25}{\sqrt{10}}\right)$