Here are summary statistics for the weights of Pepsi in randomly selected cans: $$ \mathrm{n}=36, \overrightarrow{\mathrm{x}}=0.82411 \mathrm{lb}, \mathrm{s}=0.00566 \mathrm{lb} $$. Use a confidence level of $$ 90 \% $$ to a. Identify the critical value $$ \mathrm{t}_{\alpha / 2} $$ used for finding the margin of error. $$ \mathrm{t}_{\alpha / 2}=1.69 $$ (Round to two decimal places as needed.) b. Find the margin of error. $$ \mathrm{E}=0.00159 \mathrm{lb} $$ (Round to five decimal places as needed.) c. Find the confidence interval estimate of $$ \mu $$. $$ \square \mathrm{lb}

Question

Here are summary statistics for the weights of Pepsi in randomly selected cans: $$ \mathrm{n}=36, \overrightarrow{\mathrm{x}}=0.82411 \mathrm{lb}, \mathrm{s}=0.00566 \mathrm{lb} $$. Use a confidence level of $$ 90 \% $$ to a. Identify the critical value $$ \mathrm{t}_{\alpha / 2} $$ used for finding the margin of error. $$ \mathrm{t}_{\alpha / 2}=1.69 $$ (Round to two decimal places as needed.) b. Find the margin of error. $$ \mathrm{E}=0.00159 \mathrm{lb} $$ (Round to five decimal places as needed.) c. Find the confidence interval estimate of $$ \mu $$. $$ \square \mathrm{lb}<\mu<\square \mathrm{lb} $$ (Round to five decimal places as needed.)

UpStudy AI · Accepted Answer

To solve the problem, we will follow the steps outlined in the question.

### a. Identify the critical value $$ t_{\alpha/2} $$

You have already provided the critical value for a 90% confidence level, which is:

$$
t_{\alpha/2} = 1.69
$$

### b. Find the margin of error

The margin of error $$ E $$ can be calculated using the formula:

$$
E = t_{\alpha/2} \times \frac{s}{\sqrt{n}}
$$

Where:
- $$ t_{\alpha/2} = 1.69 $$
- $$ s = 0.00566 \, \text{lb} $$
- $$ n = 36 $$

Now, we can calculate $$ \frac{s}{\sqrt{n}} $$:

$$
\frac{s}{\sqrt{n}} = \frac{0.00566}{\sqrt{36}} = \frac{0.00566}{6} = 0.00094333 \, \text{lb}
$$

Now, we can calculate the margin of error $$ E $$:

$$
E = 1.69 \times 0.00094333 \approx 0.00159 \, \text{lb}
$$

So, the margin of error is:

$$
E = 0.00159 \, \text{lb}
$$

### c. Find the confidence interval estimate of $$ \mu $$

The confidence interval for the mean $$ \mu $$ can be calculated using the formula:

$$
\overrightarrow{x} - E < \mu < \overrightarrow{x} + E
$$

Where:
- $$ \overrightarrow{x} = 0.82411 \, \text{lb} $$
- $$ E = 0.00159 \, \text{lb} $$

Calculating the lower limit:

$$
\overrightarrow{x} - E = 0.82411 - 0.00159 = 0.82252 \, \text{lb}
$$

Calculating the upper limit:

$$
\overrightarrow{x} + E = 0.82411 + 0.00159 = 0.82570 \, \text{lb}
$$

Thus, the confidence interval estimate of $$ \mu $$ is:

$$
0.82252 \, \text{lb} < \mu < 0.82570 \, \text{lb}
$$

### Final Answers

- a. $$ t_{\alpha/2} = 1.69 $$
- b. $$ E = 0.00159 \, \text{lb} $$
- c. $$ 0.82252 \, \text{lb} < \mu < 0.82570 \, \text{lb} $$
a. $$ t_{\alpha/2} = 1.69 $$ b. $$ E = 0.00159 \, \text{lb} $$ c. $$ 0.82252 \, \text{lb} < \mu < 0.82570 \, \text{lb} $$

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