ere are summary statistics for the weights of Pepsi in randomly selected​ cans: nequals36​, x overbarequals0.82408 ​lb, sequals0.00574 lb. Use a confidence level of 90​% to complete parts​ (a) and​ (b) below.
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Part 1
a. Find the confidence interval estimate of mu.
  
enter your response here lbless thanmuless than
  
enter your response here lb
​(Round to five decimal places as​ needed.)
b. Write a brief statement that interprets the confidence interval. Choose the correct answer below.
A.
One has 90​% confidence that the interval from the lower bound to the upper bound contains the true value of the population mean weight of Pepsi in a can.
B.
One has 90​% confidence that the sample mean weight of Pepsi in a can is equal to the population mean weight of Pepsi in a can.
C.
There is a 90​% chance that the true value of the population mean weight of Pepsi in a can will fall between the lower bound and the upper bound.
D.
Approximately 90​% of sample mean weights of Pepsi in a can will fall between the lower bound and the upper bound.

To find the confidence interval estimate for the mean weight of Pepsi in a can, we can use the formula for the confidence interval:

\[
\text{CI} = \bar{x} \pm z \left(\frac{s}{\sqrt{n}}\right)
\]

1. **Determine \(z\) for a 90% confidence level**: The critical value \(z\) for a 90% confidence interval is approximately 1.645.
2. **Calculate the standard error (SE)**:
   \[
   SE = \frac{s}{\sqrt{n}} = \frac{0.00574}{\sqrt{36}} = \frac{0.00574}{6} \approx 0.00095667
   \]
3. **Calculate the confidence interval**:
   \[
   \text{Margin of Error} = z \cdot SE = 1.645 \cdot 0.00095667 \approx 0.001573
   \]
   \[
   \text{Lower Bound} = \bar{x} - \text{Margin of Error} = 0.82408 - 0.001573 \approx 0.82251
   \]
   \[
   \text{Upper Bound} = \bar{x} + \text{Margin of Error} = 0.82408 + 0.001573 \approx 0.82565
   \]

Thus, the confidence interval is approximately:
\[
0.82251 \text{ lb} < \mu < 0.82565 \text{ lb}
\]

**Answer for Part 1a**:
\[
0.82251 \text{ lb} < \mu < 0.82565 \text{ lb}
\]

**Part 1b**: The correct interpretation is:
**A**. One has 90% confidence that the interval from the lower bound to the upper bound contains the true value of the population mean weight of Pepsi in a can.

Would you like more details or have any questions? Here are some follow-up questions:
1. What does the term "population mean" refer to?
2. How do you determine the critical value for different confidence levels?
3. What is the significance of the sample size in confidence intervals?
4. How would the confidence interval change with a larger sample size?
5. Why is it important to round to specific decimal places in statistics?

**Tip**: Always check if your sample is random, as this affects the validity of your confidence interval.

Here are summary statistics for the weights of Pepsi in randomly selected cans: n equals 36, x-bar equals 0.82408 lb, sequals 0.00574 lb. Use a confidence level of 90% to complete parts (a) and (b) below. Part 1 a. Find the confidence interval estimate of mu. enter your response here lb less than mu less than enter your response here lb (Round to five decimal places as needed.) b. Write a brief statement that interprets the confidence interval. Choose the correct answer below. A. One has 90% confidence that the interval from the lower bound to the upper bound contains the true value of the population mean weight of Pepsi in a can. B. One has 90% confidence that the sample mean weight of Pepsi in a can is equal to the population mean weight of Pepsi in a can. C. There is a 90% chance that the true value of the population mean weight of Pepsi in a can will fall between the lower bound and the upper bound. D. Approximately 90% of sample mean weights of Pepsi in a can will fall between the lower bound and the upper bound.

Explore how to calculate the confidence interval for the mean weight of Pepsi cans using summary statistics: n=36, x̄=0.82408 lb, and SE=0.00574 lb, with a confidence level of 90%. This resource provides key statistical concepts and detailed steps.

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