To construct a 90% confidence interval to estimate the proportion of the population that does not have the Norcovirus, let's go through the calculation step-by-step.

Step 1: Determine the sample proportion

Given:

• Total sample size, $n = 363$
• Number of people without the virus, $x = 298$

The sample proportion $\hat{p}$ is calculated as: $\hat{p} = \frac{x}{n} = \frac{298}{363} \approx 0.821$

Step 2: Determine the z-score for a 90% confidence level

For a 90% confidence level, we need the z-score that corresponds to the middle 90% of the normal distribution. This is typically $z = 1.645$.

Step 3: Calculate the standard error

The standard error (SE) of the sample proportion is calculated as: $\text{SE} = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$ Substitute $\hat{p} = 0.821$ and $n = 363$: $\text{SE} = \sqrt{\frac{0.821 \times (1 - 0.821)}{363}} \approx \sqrt{\frac{0.821 \times 0.179}{363}} \approx \sqrt{\frac{0.146859}{363}} \approx 0.0201$

Step 4: Calculate the margin of error

The margin of error (ME) is given by: $\text{ME} = z \times \text{SE} = 1.645 \times 0.0201 \approx 0.033$

Step 5: Construct the confidence interval

The confidence interval is: $\hat{p} \pm \text{ME} = 0.821 \pm 0.033$ This gives us: $\text{Lower limit} = 0.821 - 0.033 = 0.788$ $\text{Upper limit} = 0.821 + 0.033 = 0.854$

Final Answer

The 90% confidence interval to estimate the proportion of the population that does not have the Norcovirus is: $(0.788, 0.854)$

Would you like further details on any step or have additional questions?

Here are some related questions you might find helpful:

1. How do you interpret this confidence interval?
2. What would change if we used a 95% confidence level instead?
3. How would the confidence interval change with a larger sample size?
4. What is the relationship between the confidence level and the width of the interval?
5. How does the standard error impact the confidence interval?

Tip: A higher confidence level leads to a wider interval, capturing more of the population proportion range but reducing precision.