Approximately one-third of Canadians have some form of privately-covered long-term disability insurance. If random groups of 150 Canadians are sampled, find the range of values that the proportion of those with long-term disability health insurance are 80% likely to fall within.

Standard Normal Distribution Table

  < p̂ < 

Round to 3 decimal places.

To solve this, we need to apply the **Central Limit Theorem** (CLT) to estimate the range of the sample proportion that is likely to contain 80% of the data.

### Step 1: Given Information
- The proportion \( p \) of Canadians with long-term disability insurance is approximately 1/3, so:
  \[
  p = \frac{1}{3} = 0.333
  \]
- The sample size is \( n = 150 \).

### Step 2: Standard Error (SE) of the Sample Proportion
The formula for the **standard error** of the sample proportion is:
\[
SE = \sqrt{\frac{p(1 - p)}{n}}
\]
Substituting the known values:
\[
SE = \sqrt{\frac{0.333 \times (1 - 0.333)}{150}} = \sqrt{\frac{0.333 \times 0.667}{150}}
\]

### Step 3: Find the Z-values for the 80% Range
The middle 80% of the normal distribution is centered around the mean, so 10% lies in each tail. From the standard normal table, the **z-scores** for the 10th and 90th percentiles are approximately:
\[
z = \pm 1.28
\]

### Step 4: Confidence Interval Calculation
The formula for the confidence interval for the sample proportion is:
\[
p̂ = p \pm z \cdot SE
\]
Substituting the values of \( p = 0.333 \), \( z = 1.28 \), and the standard error (SE):

Let’s calculate it.The range of values for the proportion of Canadians with long-term disability insurance, in which 80% of the samples of 150 people are likely to fall, is approximately between **0.284** and **0.382**.

Would you like further details or explanations? Here are some related questions you might consider:

1. How is the sample size related to the width of the confidence interval?
2. What happens to the confidence interval if the proportion \( p \) is closer to 0 or 1?
3. How does increasing the confidence level (e.g., 90% or 95%) affect the Z-score and interval range?
4. Why do we use the normal distribution for sample proportions in this case?
5. What are other factors that could affect the interpretation of this confidence interval?

**Tip:** Larger sample sizes reduce the standard error, leading to a narrower confidence interval, assuming the same confidence level.

Approximately one-third of Canadians have some form of privately-covered long-term disability insurance. If random groups of 150 Canadians are sampled, find the range of values that the proportion of those with long-term disability health insurance are 80% likely to fall within.

Learn how to calculate the 80% confidence interval for the proportion of Canadians with privately-covered long-term disability insurance using the Central Limit Theorem. This problem uses a sample of 150 Canadians, with detailed steps explaining standard error and confidence interval formulas. Ideal for college-level statistics students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation