Question
f a test that has reliability 0.7 has the number of questions increased by a factor​ n, the reliability R of the new test is given by Upper R left parenthesis n right parenthesis equals StartFraction 0.6 n Over 0.1 plus 0.6 n EndFraction
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GPT-4o mini
The reliability of a test is a measure of its consistency or dependability. The formula you provided for the reliability, \( R(n) \), of a new test with an increased number of questions by a factor of \( n \) is:
\[
R(n) = \frac{0.6n}{0.1 + 0.6n}
\]
### Understanding the Formula
1. **Current Reliability**: The original test has a reliability of \( 0.7 \).
2. **Increased Number of Questions**: \( n \) is the factor by which the number of questions has increased.
3. **New Reliability**: The new reliability \( R(n) \) is a function of \( n \), which suggests that as the number of questions increases, the reliability of the test may also change.
### Analyzing the Formula
- **As \( n \) Increases**:
- When \( n \) gets very large, the term \( 0.1 \) in the denominator becomes negligible compared to \( 0.6n \).
- Therefore, in the limit as \( n \) approaches infinity, we can approximate the reliability:
\[
R(n) \approx \frac{0.6n}{0.6n} = 1
\]
This indicates that if you keep increasing the number of questions indefinitely, the reliability would approach 1, suggesting perfect reliability.
### Important Points:
- **Non-Negativity**: Since reliability must always be between \( 0 \) and \( 1 \), as \( n \) increases, \( R(n) \) should stay within these bounds.
- **Practical Implications**: Increasing the number of questions can potentially yield a more reliable test, but it's essential to ensure that the added questions are relevant and maintain the quality of the test.
### Conclusion:
You can use this formula to calculate the reliability for specific values of \( n \). For example, if you want to find the reliability when the number of questions has increased by a factor of 5 (i.e., \( n = 5 \)), you would plug that value into the formula:
\[
R(5) = \frac{0.6(5)}{0.1 + 0.6(5)} = \frac{3}{3.1} \approx 0.9677
\]
So, the reliability of the test with 5 times the original number of questions would be approximately 0.968.
\[
R(n) = \frac{0.6n}{0.1 + 0.6n}
\]
### Understanding the Formula
1. **Current Reliability**: The original test has a reliability of \( 0.7 \).
2. **Increased Number of Questions**: \( n \) is the factor by which the number of questions has increased.
3. **New Reliability**: The new reliability \( R(n) \) is a function of \( n \), which suggests that as the number of questions increases, the reliability of the test may also change.
### Analyzing the Formula
- **As \( n \) Increases**:
- When \( n \) gets very large, the term \( 0.1 \) in the denominator becomes negligible compared to \( 0.6n \).
- Therefore, in the limit as \( n \) approaches infinity, we can approximate the reliability:
\[
R(n) \approx \frac{0.6n}{0.6n} = 1
\]
This indicates that if you keep increasing the number of questions indefinitely, the reliability would approach 1, suggesting perfect reliability.
### Important Points:
- **Non-Negativity**: Since reliability must always be between \( 0 \) and \( 1 \), as \( n \) increases, \( R(n) \) should stay within these bounds.
- **Practical Implications**: Increasing the number of questions can potentially yield a more reliable test, but it's essential to ensure that the added questions are relevant and maintain the quality of the test.
### Conclusion:
You can use this formula to calculate the reliability for specific values of \( n \). For example, if you want to find the reliability when the number of questions has increased by a factor of 5 (i.e., \( n = 5 \)), you would plug that value into the formula:
\[
R(5) = \frac{0.6(5)}{0.1 + 0.6(5)} = \frac{3}{3.1} \approx 0.9677
\]
So, the reliability of the test with 5 times the original number of questions would be approximately 0.968.
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