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.