I need help finding the answers that go into the yellow highlighted boxes :) Thank you! ### Confidence Interval Calculation for Coffee Temperature#### Problem StatementYou randomly select 16 coffee shops and measure the temperature of the coffee sold at each. The sample mean temperature is 170.0 degrees with a sample standard deviation of 10 degrees. Construct a 99% confidence interval for the population mean temperature. Assume the temperatures are approximately normally distributed.#### Parameters Given:- **Sample mean (x-bar):** 170.0 degrees- **Sample standard deviation (std dev):** 10 degrees- **Sample size (n):** 16- **Confidence Level:** 99%#### Steps to Calculate the Confidence Interval1. **Point Estimate:** The point estimate for the population mean is the sample mean. \[ \text{Point Estimate} = \bar{x} = 170.0 \]2. **Margin of Error:** The margin of error (E) is calculated using the formula: \[ E = t_{\alpha/2} \cdot \frac{s}{\sqrt{n}} \] Where: - $ t_{\alpha/2} $ is the t-score corresponding to the 99% confidence level for degrees of freedom (df = n - 1). - $ s $ is the sample standard deviation. - $ n $ is the sample size. 3. **Lower and Upper Limits:** \[ \text{Lower Limit} = \bar{x} - E \] \[ \text{Upper Limit} = \bar{x} + E \]#### Calculations:(includes standard t-table lookup or software tools to get $ t_{\alpha/2} $)Assuming $ t_{\alpha/2} \approx 2.947 $ for a 99% confidence interval with 15 degrees of freedom,\[ E = 2.947 \cdot \frac{10}{\sqrt{16}} \approx 2.947 \cdot 2.5 = 7.3675 \]Thus,\[ \text{Lower Limit} = 170.0 - 7.4 \approx 162.6 \]\[ \text{Upper Limit} = 170.0 + 7.4 \approx 177.4 \]#### Summary:The 99% confidence interval for the population mean temperature is \([162.6, 177.4]\

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Description: I need help finding the answers that go into the yellow highlighted boxes :) Thank you! ### Confidence Interval Calculation for Coffee Temperature#### Problem StatementYou randomly select 16 coffee shops and measure the temperature of the coffee sold

Here, the temperatures are approximately normally distributed. It is appropriate to one sample z…

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Statistics

You randomly select 16 coffee shops and measure the temperature of the coffee sold at each. The sample mean temperature is 170.0 degrees with a sample standard deviation of 10 degrees. Construct a 99% confidence interval for the population mean temperature. Assume the temperatures are approximately normally distributed. x-bar std dev Confidence Level Margin of Error Point Estimate |round off to 1 decimal place round off to 1 decimal place Lower Limit Upper Limit Interpret the confidence interval in context of the problem

You randomly select 16 coffee shops and measure the temperature of the coffee sold at each. The sample mean temperature is 170.0 degrees with a sample standard deviation of 10 degrees. Construct a 99% confidence interval for the population mean temperature. Assume the temperatures are approximately normally distributed. x-bar std dev Confidence Level Margin of Error Point Estimate |round off to 1 decimal place round off to 1 decimal place Lower Limit Upper Limit Interpret the confidence interval in context of the problem

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Author:Amos Gilat

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I need help finding the answers that go into the yellow highlighted boxes :) Thank you!

### Confidence Interval Calculation for Coffee Temperature

#### Problem Statement
You randomly select 16 coffee shops and measure the temperature of the coffee sold at each. The sample mean temperature is 170.0 degrees with a sample standard deviation of 10 degrees. Construct a 99% confidence interval for the population mean temperature. Assume the temperatures are approximately normally distributed.

#### Parameters Given:
- **Sample mean (x-bar):** 170.0 degrees
- **Sample standard deviation (std dev):** 10 degrees
- **Sample size (n):** 16
- **Confidence Level:** 99%

#### Steps to Calculate the Confidence Interval

1. **Point Estimate:**
The point estimate for the population mean is the sample mean.
\[ \text{Point Estimate} = \bar{x} = 170.0 \]

2. **Margin of Error:**
The margin of error (E) is calculated using the formula:
\[ E = t_{\alpha/2} \cdot \frac{s}{\sqrt{n}} \]
Where:
- $ t_{\alpha/2} $ is the t-score corresponding to the 99% confidence level for degrees of freedom (df = n - 1).
- $ s $ is the sample standard deviation.
- $ n $ is the sample size.

3. **Lower and Upper Limits:**
\[ \text{Lower Limit} = \bar{x} - E \]
\[ \text{Upper Limit} = \bar{x} + E \]

#### Calculations:
(includes standard t-table lookup or software tools to get $ t_{\alpha/2} $)

Assuming $ t_{\alpha/2} \approx 2.947 $ for a 99% confidence interval with 15 degrees of freedom,

\[ E = 2.947 \cdot \frac{10}{\sqrt{16}} \approx 2.947 \cdot 2.5 = 7.3675 \]

Thus,

\[ \text{Lower Limit} = 170.0 - 7.4 \approx 162.6 \]
\[ \text{Upper Limit} = 170.0 + 7.4 \approx 177.4 \]

#### Summary:
The 99% confidence interval for the population mean temperature is \([162.6, 177.4]\

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