Calculate the margin of error and construct the confidence interval for the population mean (you may assume the population data is normally distributed):
Standard Normal Distribution Table
a.  x̄ =99.4,n=70, σ =1.25, α =0.1


E=
  

Round to 3 significant digits

 < μ < 
Round to 2 decimal places
b.  x̄ =51.3,n=96, σ =12.6, α =0.05


E=
  

Round to 3 significant digits

 < μ < 
Round to 2 decimal places

To solve these problems, we can use the formula for the margin of error \( E \) and the confidence interval for the population mean \( \mu \). Since we are assuming that the population is normally distributed, we can use the **Z-distribution**.

### Margin of Error Formula
\[
E = Z_{\alpha/2} \times \frac{\sigma}{\sqrt{n}}
\]
Where:
- \( Z_{\alpha/2} \) is the critical value from the standard normal distribution table.
- \( \sigma \) is the population standard deviation.
- \( n \) is the sample size.
- \( \alpha \) is the level of significance (so \( 1 - \alpha \) is the confidence level).

### Confidence Interval Formula
\[
\text{Confidence Interval} = \left( x̄ - E, x̄ + E \right)
\]
Where \( x̄ \) is the sample mean and \( E \) is the margin of error.

### Part a

Given:
- \( x̄ = 99.4 \)
- \( n = 70 \)
- \( \sigma = 1.25 \)
- \( \alpha = 0.1 \) (which gives a confidence level of \( 1 - \alpha = 90\%\))

1. **Find \( Z_{\alpha/2} \)**:
   For \( \alpha = 0.1 \), \( \alpha/2 = 0.05 \). From the standard normal distribution table, the critical value \( Z_{0.05} = 1.645 \).

2. **Calculate the margin of error** \( E \):
   \[
   E = 1.645 \times \frac{1.25}{\sqrt{70}} 
   \]

Let's compute the margin of error and the confidence interval.

### Part b

Given:
- \( x̄ = 51.3 \)
- \( n = 96 \)
- \( \sigma = 12.6 \)
- \( \alpha = 0.05 \) (which gives a confidence level of \( 1 - \alpha = 95\%\))

1. **Find \( Z_{\alpha/2} \)**:
   For \( \alpha = 0.05 \), \( \alpha/2 = 0.025 \). From the standard normal distribution table, the critical value \( Z_{0.025} = 1.96 \).

2. **Calculate the margin of error** \( E \):
   \[
   E = 1.96 \times \frac{12.6}{\sqrt{96}} 
   \]

Let's calculate these values.### Part a:
- Margin of error \( E = 0.246 \) (rounded to 3 significant digits)
- Confidence interval: \( 99.15 < \mu < 99.65 \) (rounded to 2 decimal places)

### Part b:
- Margin of error \( E = 2.52 \) (rounded to 3 significant digits)
- Confidence interval: \( 48.78 < \mu < 53.82 \) (rounded to 2 decimal places)

Would you like more details or have any questions?

Here are 5 related questions:
1. How is the critical value \( Z_{\alpha/2} \) determined for different confidence levels?
2. What is the effect of increasing the sample size on the margin of error?
3. Can the population standard deviation \( \sigma \) be estimated from the sample data?
4. How would the confidence interval change if we used a \( t \)-distribution instead of a \( Z \)-distribution?
5. What happens to the confidence interval if the significance level \( \alpha \) is reduced?

**Tip**: A smaller significance level (like 0.01 instead of 0.05) leads to a wider confidence interval since you're increasing the confidence level.

Calculate the margin of error and construct the confidence interval for the population mean (you may assume the population data is normally distributed):
a. x̄ = 99.4, n = 70, σ = 1.25, α = 0.1
b. x̄ = 51.3, n = 96, σ = 12.6, α = 0.05

Learn how to calculate the margin of error and construct confidence intervals for the population mean using sample data. The examples include a 90% confidence level for a sample mean of 99.4 (n=70, σ=1.25) and a 95% confidence level for a sample mean of 51.3 (n=96, σ=12.6).

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation