Question
Following are interest rates (annual percentage rates) for a 30-year fixed rate mortgage from a sample of lenders in Macon, Georgia for one day. It is reasonable to assume that the population is approximately normal.
4.751
4.372
4.175
4.678
4.421
4.229
4.122
4.254
3.952
4.196
construct a 95% confidence interval
4.751
4.372
4.175
4.678
4.421
4.229
4.122
4.254
3.952
4.196
construct a 95% confidence interval
Answers
GPT-4o mini
To construct a 95% confidence interval for the mean interest rate of the sample of lenders, we'll follow these steps:
1. **Calculate the sample mean (\( \bar{x} \)).**
2. **Calculate the sample standard deviation (\( s \)).**
3. **Determine the appropriate critical value from the t-distribution.**
4. **Calculate the margin of error.**
5. **Construct the confidence interval.**
### Step 1: Calculate the Sample Mean
First, we compute the mean (\( \bar{x} \)) of the interest rates.
\[
\bar{x} = \frac{\sum x_i}{n} = \frac{4.751 + 4.372 + 4.175 + 4.678 + 4.421 + 4.229 + 4.122 + 4.254 + 3.952 + 4.196}{10}
\]
Calculating the sum:
\[
4.751 + 4.372 + 4.175 + 4.678 + 4.421 + 4.229 + 4.122 + 4.254 + 3.952 + 4.196 = 43.176
\]
Now, divide by the number of observations \( n = 10 \):
\[
\bar{x} = \frac{43.176}{10} = 4.3176
\]
### Step 2: Calculate the Sample Standard Deviation
Now we calculate the standard deviation (\( s \)) using the formula:
\[
s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n - 1}}
\]
Calculating \( \sum (x_i - \bar{x})^2 \):
| \( x_i \) | \( x_i - \bar{x} \) | \( (x_i - \bar{x})^2 \) |
|------------|---------------------|--------------------------|
| 4.751 | 0.4334 | 0.1875 |
| 4.372 | 0.0544 | 0.00296 |
| 4.175 | -0.1426 | 0.0203 |
| 4.678 | 0.3604 | 0.1295 |
| 4.421 | 0.1034 | 0.0107 |
| 4.229 | -0.0886 | 0.0078 |
| 4.122 | -0.1956 | 0.0383 |
| 4.254 | -0.0636 | 0.0040 |
| 3.952 | -0.3656 | 0.1334 |
| 4.196 | -0.1216 | 0.0148 |
Now sum the squared differences:
\[
\sum (x_i - \bar{x})^2 \approx 0.1875 + 0.00296 + 0.0203 + 0.1295 + 0.0107 + 0.0078 + 0.0383 + 0.0040 + 0.1334 + 0.0148 \approx 0.5151
\]
Then plug it into the standard deviation formula:
\[
s = \sqrt{\frac{0.5151}{10 - 1}} = \sqrt{\frac{0.5151}{9}} \approx \sqrt{0.0572} \approx 0.2392
\]
### Step 3: Determine the t Critical Value
For a 95% confidence interval with \( n - 1 = 9 \) degrees of freedom, we find the t critical value. Using a t-table, the critical value for \( t_{0.025, 9} \approx 2.262 \).
### Step 4: Calculate the Margin of Error
The margin of error (ME) is calculated as follows:
\[
ME = t_{critical} \times \frac{s}{\sqrt{n}} = 2.262 \times \frac{0.2392}{\sqrt{10}} \approx 2.262 \times 0.0756 \approx 0.1716
\]
### Step 5: Construct the Confidence Interval
The confidence interval is given by:
\[
\bar{x} \pm ME
\]
Calculating the lower and upper limits:
\[
\text{Lower limit} = 4.3176 - 0.1716 \approx 4.1460
\]
\[
\text{Upper limit} = 4.3176 + 0.1716 \approx 4.4892
\]
### Final Results
Thus, the 95% confidence interval for the mean interest rate is approximately:
\[
(4.1460, 4.4892)
\]
1. **Calculate the sample mean (\( \bar{x} \)).**
2. **Calculate the sample standard deviation (\( s \)).**
3. **Determine the appropriate critical value from the t-distribution.**
4. **Calculate the margin of error.**
5. **Construct the confidence interval.**
### Step 1: Calculate the Sample Mean
First, we compute the mean (\( \bar{x} \)) of the interest rates.
\[
\bar{x} = \frac{\sum x_i}{n} = \frac{4.751 + 4.372 + 4.175 + 4.678 + 4.421 + 4.229 + 4.122 + 4.254 + 3.952 + 4.196}{10}
\]
Calculating the sum:
\[
4.751 + 4.372 + 4.175 + 4.678 + 4.421 + 4.229 + 4.122 + 4.254 + 3.952 + 4.196 = 43.176
\]
Now, divide by the number of observations \( n = 10 \):
\[
\bar{x} = \frac{43.176}{10} = 4.3176
\]
### Step 2: Calculate the Sample Standard Deviation
Now we calculate the standard deviation (\( s \)) using the formula:
\[
s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n - 1}}
\]
Calculating \( \sum (x_i - \bar{x})^2 \):
| \( x_i \) | \( x_i - \bar{x} \) | \( (x_i - \bar{x})^2 \) |
|------------|---------------------|--------------------------|
| 4.751 | 0.4334 | 0.1875 |
| 4.372 | 0.0544 | 0.00296 |
| 4.175 | -0.1426 | 0.0203 |
| 4.678 | 0.3604 | 0.1295 |
| 4.421 | 0.1034 | 0.0107 |
| 4.229 | -0.0886 | 0.0078 |
| 4.122 | -0.1956 | 0.0383 |
| 4.254 | -0.0636 | 0.0040 |
| 3.952 | -0.3656 | 0.1334 |
| 4.196 | -0.1216 | 0.0148 |
Now sum the squared differences:
\[
\sum (x_i - \bar{x})^2 \approx 0.1875 + 0.00296 + 0.0203 + 0.1295 + 0.0107 + 0.0078 + 0.0383 + 0.0040 + 0.1334 + 0.0148 \approx 0.5151
\]
Then plug it into the standard deviation formula:
\[
s = \sqrt{\frac{0.5151}{10 - 1}} = \sqrt{\frac{0.5151}{9}} \approx \sqrt{0.0572} \approx 0.2392
\]
### Step 3: Determine the t Critical Value
For a 95% confidence interval with \( n - 1 = 9 \) degrees of freedom, we find the t critical value. Using a t-table, the critical value for \( t_{0.025, 9} \approx 2.262 \).
### Step 4: Calculate the Margin of Error
The margin of error (ME) is calculated as follows:
\[
ME = t_{critical} \times \frac{s}{\sqrt{n}} = 2.262 \times \frac{0.2392}{\sqrt{10}} \approx 2.262 \times 0.0756 \approx 0.1716
\]
### Step 5: Construct the Confidence Interval
The confidence interval is given by:
\[
\bar{x} \pm ME
\]
Calculating the lower and upper limits:
\[
\text{Lower limit} = 4.3176 - 0.1716 \approx 4.1460
\]
\[
\text{Upper limit} = 4.3176 + 0.1716 \approx 4.4892
\]
### Final Results
Thus, the 95% confidence interval for the mean interest rate is approximately:
\[
(4.1460, 4.4892)
\]