here are summary statistics for the weights of pepsi in randomly selected cans: n = 36, $\bar{x}=0.82411$…

here are summary statistics for the weights of pepsi in randomly selected cans: n = 36, $\bar{x}=0.82411$ lb, s = 0.00573 lb. use a confidence level of 95% to complete parts (a) through (d) below.\na. identify the critical value $t_{alpha/2}$ used for finding the margin of error.\n$t_{alpha/2}=2.030$ (round to two decimal places as needed.)\nb. find the margin of error.\ne = $square$ lb (round to five decimal places as needed.)

here are summary statistics for the weights of pepsi in randomly selected cans: n = 36, $\bar{x}=0.82411$ lb, s = 0.00573 lb. use a confidence level of 95% to complete parts (a) through (d) below.\na. identify the critical value $t_{alpha/2}$ used for finding the margin of error.\n$t_{alpha/2}=2.030$ (round to two decimal places as needed.)\nb. find the margin of error.\ne = $square$ lb (round to five decimal places as needed.)

Answer

Explanation:

Step1: Recall margin - of - error formula

The formula for the margin of error $E$ when the population standard deviation $\sigma$ is unknown is $E = t_{\alpha/2}\frac{s}{\sqrt{n}}$, where $t_{\alpha/2}$ is the critical value, $s$ is the sample standard deviation, and $n$ is the sample size.

Step2: Identify given values

We are given that $t_{\alpha/2}=2.030$, $s = 0.00573$ lb, and $n = 36$.

Step3: Calculate the margin of error

Substitute the values into the formula: $E=2.030\times\frac{0.00573}{\sqrt{36}}$. First, $\sqrt{36}=6$. Then $\frac{0.00573}{6}= 0.000955$. Finally, $E = 2.030\times0.000955=0.00194865\approx0.00195$.

Answer:

$0.00195$