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
Unitary Method
The word “unitary” comes from the word “unit”, which means a single and complete entity. In this method, we find the value of a unit product from the given number of products, and then we solve for the other number of products.
Speed, Time, and Distance
Imagine you and 3 of your friends are planning to go to the playground at 6 in the evening. Your house is one mile away from the playground and one of your friends named Jim must start at 5 pm to reach the playground by walk. The other two friends are 3 miles away.
Profit and Loss
The amount earned or lost on the sale of one or more items is referred to as the profit or loss on that item.
Units and Measurements
Measurements and comparisons are the foundation of science and engineering. We, therefore, need rules that tell us how things are measured and compared. For these measurements and comparisons, we perform certain experiments, and we will need the experiments to set up the devices.
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![### 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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1bc805eb-60ed-457b-ab8e-02c516698c74%2F6379a949-61c5-4305-a945-71243f790d08%2F5h87qpk_processed.jpeg&w=3840&q=75)
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