Question
6. A hotel chain wants to estimate the mean number of rooms rented daily in a given month. The population of rooms rented daily is assumed to be normally distributed for each month. During February, a sample of 25 days has a sample mean of 37 rooms with a standard deviation of 24 rooms.
6.1 Find the 99% confidence interval for the mean number of rooms rented daily in a given month?
6.2 What is the 90% confidence interval calculated from the same data?
6.1 Find the 99% confidence interval for the mean number of rooms rented daily in a given month?
6.2 What is the 90% confidence interval calculated from the same data?
Answers
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Answered by
GPT-4o mini
To find the confidence intervals for the mean number of rooms rented daily, we will use the formula for the confidence interval for a mean when the population is normally distributed. Given a sample mean (\( \bar{x} \)), a standard deviation (\( s \)), and a sample size (\( n \)), the confidence interval can be calculated using the formula:
\[
\bar{x} \pm z^* \left( \frac{s}{\sqrt{n}} \right)
\]
Where:
- \( \bar{x} \) is the sample mean.
- \( z^* \) is the z-value associated with the desired confidence level.
- \( s \) is the sample standard deviation.
- \( n \) is the sample size.
### Given Data:
- Sample mean (\( \bar{x} \)) = 37 rooms
- Sample standard deviation (\( s \)) = 24 rooms
- Sample size (\( n \)) = 25 days
**6.1 Finding the 99% Confidence Interval:**
1. **Determine the z-value for 99% confidence:**
- The z-value corresponding to a 99% confidence level (from standard normal distribution tables) is approximately 2.576.
2. **Calculate the standard error (SE):**
\[
SE = \frac{s}{\sqrt{n}} = \frac{24}{\sqrt{25}} = \frac{24}{5} = 4.8
\]
3. **Calculate the confidence interval using the formula:**
\[
37 \pm 2.576 \times 4.8
\]
4. **Calculate the margin of error:**
\[
2.576 \times 4.8 \approx 12.3728
\]
5. **Confidence Interval:**
\[
37 \pm 12.3728 \Rightarrow (37 - 12.3728, 37 + 12.3728) \Rightarrow (24.6272, 49.3728)
\]
Thus, the 99% confidence interval for the mean number of rooms rented daily is approximately **(24.63, 49.37)**.
---
**6.2 Finding the 90% Confidence Interval:**
1. **Determine the z-value for 90% confidence:**
- The z-value corresponding to a 90% confidence level is approximately 1.645.
2. **Re-use the calculated standard error (SE):**
\[
SE = \frac{s}{\sqrt{n}} = 4.8
\]
3. **Calculate the confidence interval using the formula:**
\[
37 \pm 1.645 \times 4.8
\]
4. **Calculate the margin of error:**
\[
1.645 \times 4.8 \approx 7.896
\]
5. **Confidence Interval:**
\[
37 \pm 7.896 \Rightarrow (37 - 7.896, 37 + 7.896) \Rightarrow (29.104, 44.896)
\]
Thus, the 90% confidence interval for the mean number of rooms rented daily is approximately **(29.10, 44.90)**.
\[
\bar{x} \pm z^* \left( \frac{s}{\sqrt{n}} \right)
\]
Where:
- \( \bar{x} \) is the sample mean.
- \( z^* \) is the z-value associated with the desired confidence level.
- \( s \) is the sample standard deviation.
- \( n \) is the sample size.
### Given Data:
- Sample mean (\( \bar{x} \)) = 37 rooms
- Sample standard deviation (\( s \)) = 24 rooms
- Sample size (\( n \)) = 25 days
**6.1 Finding the 99% Confidence Interval:**
1. **Determine the z-value for 99% confidence:**
- The z-value corresponding to a 99% confidence level (from standard normal distribution tables) is approximately 2.576.
2. **Calculate the standard error (SE):**
\[
SE = \frac{s}{\sqrt{n}} = \frac{24}{\sqrt{25}} = \frac{24}{5} = 4.8
\]
3. **Calculate the confidence interval using the formula:**
\[
37 \pm 2.576 \times 4.8
\]
4. **Calculate the margin of error:**
\[
2.576 \times 4.8 \approx 12.3728
\]
5. **Confidence Interval:**
\[
37 \pm 12.3728 \Rightarrow (37 - 12.3728, 37 + 12.3728) \Rightarrow (24.6272, 49.3728)
\]
Thus, the 99% confidence interval for the mean number of rooms rented daily is approximately **(24.63, 49.37)**.
---
**6.2 Finding the 90% Confidence Interval:**
1. **Determine the z-value for 90% confidence:**
- The z-value corresponding to a 90% confidence level is approximately 1.645.
2. **Re-use the calculated standard error (SE):**
\[
SE = \frac{s}{\sqrt{n}} = 4.8
\]
3. **Calculate the confidence interval using the formula:**
\[
37 \pm 1.645 \times 4.8
\]
4. **Calculate the margin of error:**
\[
1.645 \times 4.8 \approx 7.896
\]
5. **Confidence Interval:**
\[
37 \pm 7.896 \Rightarrow (37 - 7.896, 37 + 7.896) \Rightarrow (29.104, 44.896)
\]
Thus, the 90% confidence interval for the mean number of rooms rented daily is approximately **(29.10, 44.90)**.
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