With interpretation please **Constructing a 95% Confidence Interval Based on Survey Data***In a survey, 25 people were asked how much they spent on their bosses' last birthday gift. The results were roughly bell-shaped with a mean of $39 and a standard deviation of $4. Construct and interpret a 95% confidence interval.***Give your answers to one decimal place:**[ ] [ ]**Interpretation:**---### Explanation:**Key points and formula:**- **Sample Size (n)** = 25- **Sample Mean (x̄)** = $39- **Sample Standard Deviation (s)** = $4- **Confidence Level** = 95%Since the sample size is 25 which is less than 30, the t-distribution is preferred over the z-distribution for constructing the confidence interval.1. **Find the t-score** associated with a 95% confidence level for 24 degrees of freedom (n-1 = 25-1 = 24). - From the t-table, for a 95% confidence level and 24 degrees of freedom, the t-score (t*) is approximately 2.064.2. **Calculate the Standard Error (SE)** of the mean: \[ \text{SE} = \frac{s}{\sqrt{n}} = \frac{4}{\sqrt{25}} = \frac{4}{5} = 0.8 \]3. **Construct the margin of error (ME)**: \[ \text{ME} = t^* \times \text{SE} = 2.064 \times 0.8 = 1.6512 \]4. **Determine the confidence interval**: - **Lower limit**: $ x̄ - \text{ME} = 39 - 1.6512 = 37.3 $ - **Upper limit**: $ x̄ + \text{ME} = 39 + 1.6512 = 40.7 $Thus, the 95% confidence interval is [37.3, 40.7].**Interpretation:**We are 95% confident that the true mean amount spent on bosses' last birthday gifts by the surveyed group lies between $37.3 and $40.7.

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Description: With interpretation please **Constructing a 95% Confidence Interval Based on Survey Data***In a survey, 25 people were asked how much they spent on their bosses' last birthday gift. The results were roughly bell-shaped with a mean of $39 and a stand

Given: n = sample size = 25 Normal distribution sample mean = $39 sample standard deviation = $4…

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In a survey, 25 people were asked how much they spent on their bosses last birthday gift. The results were roughly bell-shaped with a mean of $39 and standard deviation of $4. Construct and interpret a 95% confidence interval. Give your answers to one decimal place. Interpretation:

In a survey, 25 people were asked how much they spent on their bosses last birthday gift. The results were roughly bell-shaped with a mean of $39 and standard deviation of $4. Construct and interpret a 95% confidence interval. Give your answers to one decimal place. Interpretation:

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With interpretation please

**Constructing a 95% Confidence Interval Based on Survey Data**

*In a survey, 25 people were asked how much they spent on their bosses' last birthday gift. The results were roughly bell-shaped with a mean of $39 and a standard deviation of $4. Construct and interpret a 95% confidence interval.*

**Give your answers to one decimal place:**

[ ] [ ]

**Interpretation:**

---

### Explanation:

**Key points and formula:**
- **Sample Size (n)** = 25
- **Sample Mean (x̄)** = $39
- **Sample Standard Deviation (s)** = $4
- **Confidence Level** = 95%

Since the sample size is 25 which is less than 30, the t-distribution is preferred over the z-distribution for constructing the confidence interval.

1. **Find the t-score** associated with a 95% confidence level for 24 degrees of freedom (n-1 = 25-1 = 24).
- From the t-table, for a 95% confidence level and 24 degrees of freedom, the t-score (t*) is approximately 2.064.

2. **Calculate the Standard Error (SE)** of the mean:
\[
\text{SE} = \frac{s}{\sqrt{n}} = \frac{4}{\sqrt{25}} = \frac{4}{5} = 0.8
\]

3. **Construct the margin of error (ME)**:
\[
\text{ME} = t^* \times \text{SE} = 2.064 \times 0.8 = 1.6512
\]

4. **Determine the confidence interval**:
- **Lower limit**: $ x̄ - \text{ME} = 39 - 1.6512 = 37.3 $
- **Upper limit**: $ x̄ + \text{ME} = 39 + 1.6512 = 40.7 $

Thus, the 95% confidence interval is [37.3, 40.7].

**Interpretation:**

We are 95% confident that the true mean amount spent on bosses' last birthday gifts by the surveyed group lies between $37.3 and $40.7.

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