Did I calculate this correctly? Thank you. ### Constructing a 95% Confidence Interval for Mean Monthly Salary- **Problem Statement:** The mean monthly salary of a random sample of 25 college graduates under the age of 35 was found to be $2425 with a standard deviation of $635. Assume that the distribution of monthly salaries for all college graduates under the age of 35 is approximately normally distributed. We need to construct a 95% confidence interval for the population mean of monthly salaries of all college graduates under the age of 35.#### Given Data:- Sample size ($ n $): 25- Sample mean ($ \overline{X} $): $2425- Standard deviation ($ s $): $635#### Step-by-Step Solution:1. **Determine the degrees of freedom (d.f.):** \[ \text{d.f.} = n - 1 = 25 - 1 = 24 \]2. **Find the critical value ($ t_c $) for a 95% confidence interval** - From the $ t $-distribution table, for 24 degrees of freedom and a 95% confidence level, ($ \alpha/2 = 0.025 $), the critical value ($ t_c $) is approximately 2.064.3. **Applying the formula for the confidence interval:** \[ \overline{X} \pm t_c \left( \frac{s}{\sqrt{n}} \right) \]4. **Plugging in the values:** \[ 2425 \pm 2.064 \left( \frac{635}{\sqrt{25}} \right) \] \[ 2425 \pm 2.064 \left( \frac{635}{5} \right) \] \[ 2425 \pm 2.064 \times 127 \] \[ 2425 \pm 262.128 \]5. **Calculate the intervals:** \[ \text{Lower bound} = 2425 - 262.128 = 2162.872 \] \[ \text{Upper bound} = 2425 + 262.128 = 2687.128 \]#### Conclusion:The 95% confidence interval for

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Description: Did I calculate this correctly? Thank you. ### Constructing a 95% Confidence Interval for Mean Monthly Salary- **Problem Statement:** The mean monthly salary of a random sample of 25 college graduates under the age of 35 was found to be $2425 with a

We have given that Sample size n=25 , sample mean xbar=2425 , and standard deviation s=635 tc=2.064…

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7. The mean monthly salary of a random sample of 25 college graduates under the age of 35 was found to be $2425 with a standard deviation of $635 Assume that the distribution of monthly salaries for all college graduates under the age of 35 is approximately normally distributed. Construct a 95% confidence interval for u, the population mean of monthly salaries of all college graduates under the 35. age of n= as &=2425 S= 635 d.f= n-1 = 25-1 d.f.= 24 9520 C.I = 1.96 ヌキt。% 2=.25 950 /2:.25 2425+20/635 Vas t=2064 287, 2687.13

7. The mean monthly salary of a random sample of 25 college graduates under the age of 35 was found to be $2425 with a standard deviation of $635 Assume that the distribution of monthly salaries for all college graduates under the age of 35 is approximately normally distributed. Construct a 95% confidence interval for u, the population mean of monthly salaries of all college graduates under the 35. age of n= as &=2425 S= 635 d.f= n-1 = 25-1 d.f.= 24 9520 C.I = 1.96 ヌキt。% 2=.25 950 /2:.25 2425+20/635 Vas t=2064 287, 2687.13

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Concept explainers

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Did I calculate this correctly? Thank you.

### Constructing a 95% Confidence Interval for Mean Monthly Salary

- **Problem Statement:**
The mean monthly salary of a random sample of 25 college graduates under the age of 35 was found to be $2425 with a standard deviation of $635. Assume that the distribution of monthly salaries for all college graduates under the age of 35 is approximately normally distributed. We need to construct a 95% confidence interval for the population mean of monthly salaries of all college graduates under the age of 35.

#### Given Data:
- Sample size ($ n $): 25
- Sample mean ($ \overline{X} $): $2425
- Standard deviation ($ s $): $635

#### Step-by-Step Solution:

1. **Determine the degrees of freedom (d.f.):**
\[
\text{d.f.} = n - 1 = 25 - 1 = 24
\]

2. **Find the critical value ($ t_c $) for a 95% confidence interval**
- From the $ t $-distribution table, for 24 degrees of freedom and a 95% confidence level, ($ \alpha/2 = 0.025 $), the critical value ($ t_c $) is approximately 2.064.

3. **Applying the formula for the confidence interval:**
\[
\overline{X} \pm t_c \left( \frac{s}{\sqrt{n}} \right)
\]

4. **Plugging in the values:**
\[
2425 \pm 2.064 \left( \frac{635}{\sqrt{25}} \right)
\]
\[
2425 \pm 2.064 \left( \frac{635}{5} \right)
\]
\[
2425 \pm 2.064 \times 127
\]
\[
2425 \pm 262.128
\]

5. **Calculate the intervals:**
\[
\text{Lower bound} = 2425 - 262.128 = 2162.872
\]
\[
\text{Upper bound} = 2425 + 262.128 = 2687.128
\]

#### Conclusion:
The 95% confidence interval for

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