Math Problem Statement

Consider a normal population distribution with the value of 𝜎 known.

(a)

What is the confidence level for the interval 

x ± 2.88𝜎⁄

n

?

 (Round your answer to one decimal place.)

 %

(b)

What is the confidence level for the interval 

x ± 1.41𝜎⁄

n

?

 (Round your answer to one decimal place.)

 %

(c)

What value of z𝛼⁄2 in the CI formula below results in a confidence level of 99.7%? (Round your answer to two decimal places.)

 x − z𝛼⁄2 · 

𝜎

n

x + z𝛼⁄2 · 

𝜎

n   z𝛼⁄2 = 

(d)

Answer the question posed in part (c) for a confidence level of 62%. (Round your answer to two decimal places.)

z𝛼⁄2 = 

Solution

To solve this, we need to understand the relationships between the z-scores and their associated confidence levels in a normal distribution. Let’s break it down step by step.


(a) Confidence level for x±2.88σnx \pm \frac{2.88\sigma}{\sqrt{n}}

  • The z-score is z=2.88z = 2.88.
  • Confidence level corresponds to the area under the normal curve between z-z and zz. This is calculated as:

Confidence Level=Cumulative Area(2.88Z2.88)\text{Confidence Level} = \text{Cumulative Area}(-2.88 \leq Z \leq 2.88)

Using the standard normal table or a calculator:

Cumulative Probability at z=2.88=0.9980\text{Cumulative Probability at } z = 2.88 = 0.9980

The confidence level is:

Confidence Level=0.9980(10.9980)=99.6%\text{Confidence Level} = 0.9980 - (1 - 0.9980) = 99.6\%

Answer: 99.6%99.6\%


(b) Confidence level for x±1.41σnx \pm \frac{1.41\sigma}{\sqrt{n}}

  • The z-score is z=1.41z = 1.41.
  • Confidence level corresponds to the cumulative area under the normal curve between 1.41-1.41 and 1.411.41.

Using the standard normal table or a calculator:

Cumulative Probability at z=1.41=0.9207\text{Cumulative Probability at } z = 1.41 = 0.9207

The confidence level is:

Confidence Level=0.9207(10.9207)=92.1%\text{Confidence Level} = 0.9207 - (1 - 0.9207) = 92.1\%

Answer: 92.1%92.1\%


(c) What value of zα/2z_{\alpha/2} results in a 99.7% confidence level?

The confidence level is 99.7%, so the total area outside this range is:

α=10.997=0.003\alpha = 1 - 0.997 = 0.003

Each tail will have:

α/2=0.0015\alpha/2 = 0.0015

We look up the z-score for the cumulative probability of 10.0015=0.99851 - 0.0015 = 0.9985 in the standard normal table. This corresponds to:

zα/2=3.00z_{\alpha/2} = 3.00

Answer: 3.003.00


(d) What value of zα/2z_{\alpha/2} results in a 62% confidence level?

The confidence level is 62%, so the total area outside this range is:

α=10.62=0.38\alpha = 1 - 0.62 = 0.38

Each tail will have:

α/2=0.19\alpha/2 = 0.19

We look up the z-score for the cumulative probability of 10.19=0.811 - 0.19 = 0.81 in the standard normal table. This corresponds to:

zα/2=0.88z_{\alpha/2} = 0.88

Answer: 0.880.88


Final Answers:

(a) 99.6%99.6\%
(b) 92.1%92.1\%
(c) 3.003.00
(d) 0.880.88


Would you like me to explain any step in more detail? Here are related questions you might consider:

  1. How do we interpret z-scores in confidence intervals?
  2. How do you calculate the cumulative area for non-standard z-scores?
  3. How do confidence intervals change with increasing zz-values?
  4. What is the relationship between confidence levels and significance levels?
  5. Why do we divide α\alpha by 2 in confidence interval calculations?

Tip: Always ensure your cumulative probabilities correspond to the desired confidence level to avoid calculation errors.

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Math Problem Analysis

Mathematical Concepts

Normal Distribution
Confidence Intervals
z-scores
Cumulative Probability

Formulas

Confidence Interval Formula: x ± z𝛼/2 · σ/√n
z-score lookup in cumulative probability tables
α = 1 - Confidence Level

Theorems

Properties of the Normal Distribution

Suitable Grade Level

Undergraduate (Statistics or Introductory Probability)