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 (d) Answer the question posed in part (c) for a confidence level of 70%. (Round your answer to two decimal places.)

Solution

Let's work through each part of the question methodically.

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

The general form for a confidence interval is:

x±zα/2σnx \pm z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}}

Where:

  • zα/2z_{\alpha/2} is the critical value from the standard normal distribution that corresponds to the desired confidence level.
  • σ\sigma is the known population standard deviation.
  • nn is the sample size.

In this case, we are given zα/2=2.88z_{\alpha/2} = 2.88.

To find the confidence level, we can use the relationship between zα/2z_{\alpha/2} and the confidence level. The area between zα/2-z_{\alpha/2} and +zα/2+z_{\alpha/2} represents the proportion of the standard normal distribution covered by the confidence interval.

Using a standard normal distribution table or calculator, the area to the left of z=2.88z = 2.88 is approximately 0.9980. The total area covered by the confidence interval is therefore:

2×(0.99800.5)=0.9962 \times (0.9980 - 0.5) = 0.996

So, the confidence level is approximately 99.6%.

Answer: The confidence level is 99.6%.

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

Given zα/2=1.41z_{\alpha/2} = 1.41, we follow the same procedure as in part (a).

The area to the left of z=1.41z = 1.41 is approximately 0.9207. The total area covered by the confidence interval is:

2×(0.92070.5)=0.84142 \times (0.9207 - 0.5) = 0.8414

So, the confidence level is approximately 84.1%.

Answer: The confidence level is 84.1%.

(c) Value of zα/2z_{\alpha/2} for a 99.7% confidence level

For a 99.7% confidence level, the area in the tails is:

10.997=0.0031 - 0.997 = 0.003

Dividing this evenly between the two tails gives:

α/2=0.0015\alpha/2 = 0.0015

Looking up the corresponding zα/2z_{\alpha/2} value for α/2=0.0015\alpha/2 = 0.0015 in the standard normal distribution table gives zα/2=3.00z_{\alpha/2} = 3.00.

Answer: The value of zα/2z_{\alpha/2} for a 99.7% confidence level is 3.00.

(d) Value of zα/2z_{\alpha/2} for a 70% confidence level

For a 70% confidence level, the area in the tails is:

10.70=0.301 - 0.70 = 0.30

Dividing this evenly between the two tails gives:

α/2=0.15\alpha/2 = 0.15

Looking up the corresponding zα/2z_{\alpha/2} value for α/2=0.15\alpha/2 = 0.15 in the standard normal distribution table gives zα/2=1.04z_{\alpha/2} = 1.04.

Answer: The value of zα/2z_{\alpha/2} for a 70% confidence level is 1.04.


Let me know if you want further clarification! Here are some relative questions for practice:

  1. How is the confidence interval affected if the sample size increases?
  2. What happens to the width of the confidence interval if the confidence level increases?
  3. How do you interpret a 95% confidence interval in practice?
  4. Why is the standard normal distribution used for constructing confidence intervals?
  5. How does the z-value change for a 90% confidence level?

Tip: As the confidence level increases, the corresponding z-value also increases, making the confidence interval wider.

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Math Problem Analysis

Mathematical Concepts

Confidence Interval
Standard Normal Distribution
Z-Scores

Formulas

Confidence Interval: x ± z_{α/2} * (σ/√n)
Z-Score formula: z = (X - μ) / σ

Theorems

Standard Normal Distribution Theorem
Empirical Rule (68-95-99.7 Rule)

Suitable Grade Level

College Level