A 100(1a) % confidence interval for the mean of a normal population when the value of o is known is given by (x-²α/2 By how much must the sample size n be increased if the width of the CI above is to be halved? Halving the length requires n to be increased by a factor of [ If the sample size is increased by a factor of 25, what effect will this have on the width of the interval? Increasing the sample size by a factor of 25 will ---Select--- the length by a factor of ₂X+Za/²2 ) x+ Ma
A 100(1a) % confidence interval for the mean of a normal population when the value of o is known is given by (x-²α/2 By how much must the sample size n be increased if the width of the CI above is to be halved? Halving the length requires n to be increased by a factor of [ If the sample size is increased by a factor of 25, what effect will this have on the width of the interval? Increasing the sample size by a factor of 25 will ---Select--- the length by a factor of ₂X+Za/²2 ) x+ Ma
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter13: Probability And Calculus
Section13.3: Special Probability Density Functions
Problem 15E: Describe the standard normal distribution. What are its characteristics?
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![**Understanding Confidence Intervals for Normal Distributions**
**Concept**
A 100(1 − α)% confidence interval for the mean \( \mu \) of a normal population, when the value of \( \sigma \) is known, is given by:
\[
\left( \bar{x} - z_{\alpha / 2} \cdot \frac{\sigma}{\sqrt{n}} , \; \bar{x} + z_{\alpha / 2} \cdot \frac{\sigma}{\sqrt{n}} \right).
\]
Here:
- \( \bar{x} \) is the sample mean,
- \( z_{\alpha / 2} \) is the z-score corresponding to the desired confidence level,
- \( \sigma \) is the population standard deviation,
- \( n \) is the sample size.
**Key Questions**
1. **How much must the sample size \( n \) be increased if the width of the confidence interval (CI) above is to be halved?**
Halving the interval length requires \( n \) to be increased by a factor of **4**.
2. **What effect will increasing the sample size by a factor of 25 have on the width of the interval?**
Increasing the sample size by a factor of 25 will **decrease** the length by a factor of **5**.
**Explanation**
The width of the confidence interval depends on the term \( \frac{\sigma}{\sqrt{n}} \). If the sample size \( n \) increases, the denominator of this term increases, reducing the overall value and thus narrowing the interval.
- To halve the interval width: \( \frac{1}{\sqrt{n}} \) must be halved, so \( n \) needs to be increased by a factor of \(2^2 = 4\).
- If the sample size is increased by a factor of 25: the interval width will decrease by a factor of \( \sqrt{25} = 5 \).
This understanding is crucial for designing experiments and surveys to ensure precision in estimates. Always consider the trade-off between increasing sample size and the resulting narrower confidence interval.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffb989860-4d1c-4fb5-9e6b-42a4528dce9c%2F3a96bbbb-6be5-4786-a2af-5be49c1d0f4a%2Fws8kzc_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Understanding Confidence Intervals for Normal Distributions**
**Concept**
A 100(1 − α)% confidence interval for the mean \( \mu \) of a normal population, when the value of \( \sigma \) is known, is given by:
\[
\left( \bar{x} - z_{\alpha / 2} \cdot \frac{\sigma}{\sqrt{n}} , \; \bar{x} + z_{\alpha / 2} \cdot \frac{\sigma}{\sqrt{n}} \right).
\]
Here:
- \( \bar{x} \) is the sample mean,
- \( z_{\alpha / 2} \) is the z-score corresponding to the desired confidence level,
- \( \sigma \) is the population standard deviation,
- \( n \) is the sample size.
**Key Questions**
1. **How much must the sample size \( n \) be increased if the width of the confidence interval (CI) above is to be halved?**
Halving the interval length requires \( n \) to be increased by a factor of **4**.
2. **What effect will increasing the sample size by a factor of 25 have on the width of the interval?**
Increasing the sample size by a factor of 25 will **decrease** the length by a factor of **5**.
**Explanation**
The width of the confidence interval depends on the term \( \frac{\sigma}{\sqrt{n}} \). If the sample size \( n \) increases, the denominator of this term increases, reducing the overall value and thus narrowing the interval.
- To halve the interval width: \( \frac{1}{\sqrt{n}} \) must be halved, so \( n \) needs to be increased by a factor of \(2^2 = 4\).
- If the sample size is increased by a factor of 25: the interval width will decrease by a factor of \( \sqrt{25} = 5 \).
This understanding is crucial for designing experiments and surveys to ensure precision in estimates. Always consider the trade-off between increasing sample size and the resulting narrower confidence interval.
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