Salaries of 48 college graduates who took a statistics course in college have a mean, x, of $64,700. Assuming a standard deviation, o, of $18,673, construct a 90% confidence interval for estimating the population mean μ. Click here to view a t distribution table. Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table. $<μ<$ (Round to the nearest integer as needed.)
Salaries of 48 college graduates who took a statistics course in college have a mean, x, of $64,700. Assuming a standard deviation, o, of $18,673, construct a 90% confidence interval for estimating the population mean μ. Click here to view a t distribution table. Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table. $<μ<$ (Round to the nearest integer as needed.)
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
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![### Constructing a 90% Confidence Interval for Population Mean
Given:
- Sample size (\(n\)): 48 college graduates
- Sample mean (\(\bar{x}\)): $64,700
- Standard deviation (\(\sigma\)): $18,673
We aim to construct a 90% confidence interval for the population mean (\(\mu\)).
#### Steps to Construct the Confidence Interval
1. **Identify the z-score:** For a 90% confidence level, we need to find the z-score that corresponds to the middle 90% of the standard normal distribution. The z-score for a 90% confidence interval is approximately 1.645 (since 5% is in each tail of the normal distribution).
2. **Calculate the Standard Error (SE):**
\[ SE = \frac{\sigma}{\sqrt{n}} = \frac{18,673}{\sqrt{48}} \]
3. **Margin of Error (ME):**
\[ ME = z \times SE \]
\[ ME = 1.645 \times \frac{18,673}{\sqrt{48}} \]
4. **Confidence Interval:**
\[ \bar{x} \pm ME \]
The resulting interval will provide the range in which the population mean (\(\mu\)) is expected to lie with 90% confidence.
#### Resources
- [View a t-distribution table](#)
- [View page 1 of the standard normal distribution table](#)
- [View page 2 of the standard normal distribution table](#)
#### Solution
\[ \$ \ \_ \_\_ \ \ \ < \mu < \$ \ \_ \_\_ \ (Round to the nearest integer as needed.) \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F49a5c0da-d10b-4c66-9f47-a76740d10419%2Fdbef611a-ddbe-4d4e-9f7a-79edc0b8de5f%2Fbmb5okv_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Constructing a 90% Confidence Interval for Population Mean
Given:
- Sample size (\(n\)): 48 college graduates
- Sample mean (\(\bar{x}\)): $64,700
- Standard deviation (\(\sigma\)): $18,673
We aim to construct a 90% confidence interval for the population mean (\(\mu\)).
#### Steps to Construct the Confidence Interval
1. **Identify the z-score:** For a 90% confidence level, we need to find the z-score that corresponds to the middle 90% of the standard normal distribution. The z-score for a 90% confidence interval is approximately 1.645 (since 5% is in each tail of the normal distribution).
2. **Calculate the Standard Error (SE):**
\[ SE = \frac{\sigma}{\sqrt{n}} = \frac{18,673}{\sqrt{48}} \]
3. **Margin of Error (ME):**
\[ ME = z \times SE \]
\[ ME = 1.645 \times \frac{18,673}{\sqrt{48}} \]
4. **Confidence Interval:**
\[ \bar{x} \pm ME \]
The resulting interval will provide the range in which the population mean (\(\mu\)) is expected to lie with 90% confidence.
#### Resources
- [View a t-distribution table](#)
- [View page 1 of the standard normal distribution table](#)
- [View page 2 of the standard normal distribution table](#)
#### Solution
\[ \$ \ \_ \_\_ \ \ \ < \mu < \$ \ \_ \_\_ \ (Round to the nearest integer as needed.) \]
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