Question
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
GPT-4o mini
To calculate confidence intervals for the mean number of rooms rented daily, we can use the formula for the confidence interval for the mean when the population standard deviation is unknown, which follows a t-distribution. The formula is:
\[
\text{CI} = \bar{x} \pm t^* \left( \frac{s}{\sqrt{n}} \right)
\]
where:
- \(\bar{x}\) is the sample mean,
- \(t^*\) is the critical t-value for the desired confidence level,
- \(s\) is the sample standard deviation,
- \(n\) is the sample size.
Given data:
- Sample size (\(n\)) = 25
- Sample mean (\(\bar{x}\)) = 37 rooms
- Sample standard deviation (\(s\)) = 24 rooms
### 6.1: 99% Confidence Interval
1. **Determine the critical t-value**:
For a 99% confidence level and \(n - 1 = 25 - 1 = 24\) degrees of freedom, we need to find the t-value that corresponds to \(\alpha/2 = 0.005\) (since 99% confidence means 1% in the tail).
You can find the critical value \(t_{0.005, 24}\) in a t-distribution table or using statistical software/calculator. The critical t-value for 24 degrees of freedom at the 99% confidence level is approximately:
\[
t^* \approx 2.797
\]
2. **Calculate the standard error**:
\[
\text{SE} = \frac{s}{\sqrt{n}} = \frac{24}{\sqrt{25}} = \frac{24}{5} = 4.8
\]
3. **Calculate the confidence interval**:
\[
\text{CI} = 37 \pm 2.797 \times 4.8
\]
\[
\text{CI} = 37 \pm 13.4376
\]
\[
\text{CI} = (37 - 13.4376, 37 + 13.4376) = (23.5624, 50.4376)
\]
So, the **99% confidence interval** is **(23.56, 50.44)** (rounded to two decimal places).
### 6.2: 90% Confidence Interval
1. **Determine the critical t-value**:
For a 90% confidence level and \(n - 1 = 24\) degrees of freedom, we find the t-value that corresponds to \(\alpha/2 = 0.05\).
The critical t-value for 24 degrees of freedom at the 90% confidence level is approximately:
\[
t^* \approx 1.711
\]
2. **Calculate the confidence interval**:
\[
\text{CI} = 37 \pm 1.711 \times 4.8
\]
\[
\text{CI} = 37 \pm 8.2056
\]
\[
\text{CI} = (37 - 8.2056, 37 + 8.2056) = (28.7944, 45.2056)
\]
So, the **90% confidence interval** is **(28.79, 45.21)** (rounded to two decimal places).
### Summary
- **99% Confidence Interval:** \( (23.56, 50.44) \)
- **90% Confidence Interval:** \( (28.79, 45.21) \)
\[
\text{CI} = \bar{x} \pm t^* \left( \frac{s}{\sqrt{n}} \right)
\]
where:
- \(\bar{x}\) is the sample mean,
- \(t^*\) is the critical t-value for the desired confidence level,
- \(s\) is the sample standard deviation,
- \(n\) is the sample size.
Given data:
- Sample size (\(n\)) = 25
- Sample mean (\(\bar{x}\)) = 37 rooms
- Sample standard deviation (\(s\)) = 24 rooms
### 6.1: 99% Confidence Interval
1. **Determine the critical t-value**:
For a 99% confidence level and \(n - 1 = 25 - 1 = 24\) degrees of freedom, we need to find the t-value that corresponds to \(\alpha/2 = 0.005\) (since 99% confidence means 1% in the tail).
You can find the critical value \(t_{0.005, 24}\) in a t-distribution table or using statistical software/calculator. The critical t-value for 24 degrees of freedom at the 99% confidence level is approximately:
\[
t^* \approx 2.797
\]
2. **Calculate the standard error**:
\[
\text{SE} = \frac{s}{\sqrt{n}} = \frac{24}{\sqrt{25}} = \frac{24}{5} = 4.8
\]
3. **Calculate the confidence interval**:
\[
\text{CI} = 37 \pm 2.797 \times 4.8
\]
\[
\text{CI} = 37 \pm 13.4376
\]
\[
\text{CI} = (37 - 13.4376, 37 + 13.4376) = (23.5624, 50.4376)
\]
So, the **99% confidence interval** is **(23.56, 50.44)** (rounded to two decimal places).
### 6.2: 90% Confidence Interval
1. **Determine the critical t-value**:
For a 90% confidence level and \(n - 1 = 24\) degrees of freedom, we find the t-value that corresponds to \(\alpha/2 = 0.05\).
The critical t-value for 24 degrees of freedom at the 90% confidence level is approximately:
\[
t^* \approx 1.711
\]
2. **Calculate the confidence interval**:
\[
\text{CI} = 37 \pm 1.711 \times 4.8
\]
\[
\text{CI} = 37 \pm 8.2056
\]
\[
\text{CI} = (37 - 8.2056, 37 + 8.2056) = (28.7944, 45.2056)
\]
So, the **90% confidence interval** is **(28.79, 45.21)** (rounded to two decimal places).
### Summary
- **99% Confidence Interval:** \( (23.56, 50.44) \)
- **90% Confidence Interval:** \( (28.79, 45.21) \)