Follow PANIC acronym ### Confidence Interval Estimation for Population MeanA NAPA Auto Parts supplier wants information about how long car owners plan to keep their cars. A random sample of 35 car owners results in a sample average of 7.01 years and a standard deviation of 3.74 years. Assume that the sample is drawn from a normally distributed population.Find a 95% confidence interval estimate of the population mean.### Step-by-Step Approach: PANIC#### **P – PARAMETER**Describe the parameter in context. This should be a complete sentence.- **Context**: The parameter is the average number of years that car owners plan to keep their cars. #### **A – ASSUMPTIONS**Are we allowed to construct an interval and why?- **Justification**: We can construct a confidence interval since the sample size is 35, which is greater than 30, making the Central Limit Theorem applicable. Additionally, it is given that the sample is drawn from a normally distributed population.#### **N – NAME THE INTERVAL**State what command you will be using to compute the interval along with the values you need.- **Command**: We will use the formula for a confidence interval for the population mean:\[ \bar{x} \pm t^* \left(\frac{s}{\sqrt{n}}\right) \]where $ \bar{x} $ is the sample mean, $ s $ is the sample standard deviation, $ n $ is the sample size, and $ t^* $ is the critical value from the t-distribution for 95% confidence.#### **I – INTERVAL CALCULATION**Find the interval.1. **Sample Mean ($ \bar{x} $)**: 7.01 years2. **Sample Standard Deviation ($ s $)**: 3.74 years3. **Sample Size ($ n $)**: 354. **Degrees of Freedom (df)**: $ n - 1 = 34 $5. **Critical Value ($ t^* $)**: The critical t-value for 34 degrees of freedom at a 95% confidence level can be found using a t-table or calculator, approximately 2.032.Using the formula:\[ 7.01 \pm 2.032 \left(\frac{3.74}{\sqrt{35}}\right) \]

Name: Follow PANIC acronym ### Confidence Interval Estimation for Population
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Description: Follow PANIC acronym ### Confidence Interval Estimation for Population MeanA NAPA Auto Parts supplier wants information about how long car owners plan to keep their cars. A random sample of 35 car owners results in a sample average of 7.01 years and

Confidence interval gives a range of values for the unknown parameter of the population. The width…

Math

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A NAPA Auto Parts supplier wants information about how long car owner plan to keep their cars. A random sample of 35 car owners results in a sample average of 7.01 years and a standard deviations of 3.74 years. Assume that the sample is drawn from a normally distributed population. Find a 95% confidence interval estimate of the population mean • P- PARAMETER o Describe the parameter in context. This should be a complete sentence • A- ASSUMPTIONS o Are we allowed to construct an interval and why? • N- NAME THE INTERVAL o State what command you will be using the compute the interval along with the values you need. • 1 - INTERVAL CALCULATION o Find the interval C- CONCLUSION

A NAPA Auto Parts supplier wants information about how long car owner plan to keep their cars. A random sample of 35 car owners results in a sample average of 7.01 years and a standard deviations of 3.74 years. Assume that the sample is drawn from a normally distributed population. Find a 95% confidence interval estimate of the population mean • P- PARAMETER o Describe the parameter in context. This should be a complete sentence • A- ASSUMPTIONS o Are we allowed to construct an interval and why? • N- NAME THE INTERVAL o State what command you will be using the compute the interval along with the values you need. • 1 - INTERVAL CALCULATION o Find the interval C- CONCLUSION

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Concept explainers

Contingency Table

A contingency table can be defined as the visual representation of the relationship between two or more categorical variables that can be evaluated and registered. It is a categorical version of the scatterplot, which is used to investigate the linear relationship between two variables. A contingency table is indeed a type of frequency distribution table that displays two variables at the same time.

Binomial Distribution

Binomial is an algebraic expression of the sum or the difference of two terms. Before knowing about binomial distribution, we must know about the binomial theorem.

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Question

Follow PANIC acronym

### Confidence Interval Estimation for Population Mean

A NAPA Auto Parts supplier wants information about how long car owners plan to keep their cars. A random sample of 35 car owners results in a sample average of 7.01 years and a standard deviation of 3.74 years. Assume that the sample is drawn from a normally distributed population.

Find a 95% confidence interval estimate of the population mean.

### Step-by-Step Approach: PANIC

#### **P – PARAMETER**
Describe the parameter in context. This should be a complete sentence.
- **Context**: The parameter is the average number of years that car owners plan to keep their cars.

#### **A – ASSUMPTIONS**
Are we allowed to construct an interval and why?
- **Justification**: We can construct a confidence interval since the sample size is 35, which is greater than 30, making the Central Limit Theorem applicable. Additionally, it is given that the sample is drawn from a normally distributed population.

#### **N – NAME THE INTERVAL**
State what command you will be using to compute the interval along with the values you need.
- **Command**: We will use the formula for a confidence interval for the population mean:
\[ \bar{x} \pm t^* \left(\frac{s}{\sqrt{n}}\right) \]
where $ \bar{x} $ is the sample mean, $ s $ is the sample standard deviation, $ n $ is the sample size, and $ t^* $ is the critical value from the t-distribution for 95% confidence.

#### **I – INTERVAL CALCULATION**
Find the interval.
1. **Sample Mean ($ \bar{x} $)**: 7.01 years
2. **Sample Standard Deviation ($ s $)**: 3.74 years
3. **Sample Size ($ n $)**: 35
4. **Degrees of Freedom (df)**: $ n - 1 = 34 $
5. **Critical Value ($ t^* $)**: The critical t-value for 34 degrees of freedom at a 95% confidence level can be found using a t-table or calculator, approximately 2.032.

Using the formula:
\[ 7.01 \pm 2.032 \left(\frac{3.74}{\sqrt{35}}\right) \]

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