In the spring of 2017, the Consumer Reports National Research Center conducted a survey of 1,007 adults to learn about their major healthcare concerns. The survey results showed that 577 of the respondents lack confidence they will be able to afford health insurance in the future.
(a)
What is the point estimate of the population proportion of adults who lack confidence they will be able to afford health insurance in the future. (Round your answer to two decimal places.)
(b)
At 90% confidence, what is the margin of error? (Round your answer to four decimal places.)
(c)
Develop a 90% confidence interval for the population proportion of adults who lack confidence they will be able to afford health insurance in the future. (Round your answer to four decimal places.)
to
(d)
Develop a 95% confidence interval for this population proportion. (Round your answer to four decimal places.)

Let's work through each part of this problem step by step.

### Part (a) - Point Estimate

The point estimate for the population proportion \( \hat{p} \) is calculated as:

\[
\hat{p} = \frac{\text{Number of respondents who lack confidence}}{\text{Total number of respondents}} = \frac{577}{1007}
\]

Calculating this:

\[
\hat{p} = \frac{577}{1007} \approx 0.573
\]

So, the point estimate of the population proportion is \( \hat{p} = 0.57 \) (rounded to two decimal places).

---

### Part (b) - Margin of Error (90% Confidence Level)

The formula for the margin of error (ME) at a confidence level is:

\[
\text{ME} = z_{\alpha/2} \times \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}
\]

Where:
- \( z_{\alpha/2} \) is the critical value for the given confidence level,
- \( \hat{p} \) is the sample proportion (calculated in part (a)),
- \( n \) is the sample size.

For a 90% confidence level, \( z_{\alpha/2} \) (the z-value for 90% confidence) is approximately 1.645.

Using the values:
- \( \hat{p} = 0.573 \),
- \( n = 1007 \),
- \( z_{\alpha/2} = 1.645 \),

The margin of error is:

\[
\text{ME} = 1.645 \times \sqrt{\frac{0.573(1 - 0.573)}{1007}} \approx 1.645 \times 0.0157 \approx 0.0258
\]

So, the margin of error at 90% confidence is approximately \( \text{ME} = 0.0258 \) (rounded to four decimal places).

---

### Part (c) - 90% Confidence Interval

The confidence interval (CI) is calculated using:

\[
\text{CI} = \hat{p} \pm \text{ME}
\]

Using \( \hat{p} = 0.573 \) and \( \text{ME} = 0.0258 \):

\[
\text{Lower bound} = 0.573 - 0.0258 = 0.5472
\]
\[
\text{Upper bound} = 0.573 + 0.0258 = 0.5988
\]

Thus, the 90% confidence interval is:

\[
(0.5472, 0.5988)
\]

---

### Part (d) - 95% Confidence Interval

For a 95% confidence level, the critical value \( z_{\alpha/2} \) is approximately 1.96.

The margin of error for the 95% confidence level is:

\[
\text{ME} = 1.96 \times \sqrt{\frac{0.573(1 - 0.573)}{1007}} \approx 1.96 \times 0.0157 \approx 0.0308
\]

Using this margin of error, the 95% confidence interval is:

\[
\text{Lower bound} = 0.573 - 0.0308 = 0.5422
\]
\[
\text{Upper bound} = 0.573 + 0.0308 = 0.6038
\]

Thus, the 95% confidence interval is:

\[
(0.5422, 0.6038)
\]

---

### Summary of Results
(a) Point estimate: \( \hat{p} = 0.57 \)  
(b) Margin of error at 90% confidence: \( \text{ME} = 0.0258 \)  
(c) 90% Confidence interval: \( (0.5472, 0.5988) \)  
(d) 95% Confidence interval: \( (0.5422, 0.6038) \)

Would you like any more details or clarification on any step?

Here are some related questions you might find useful:
1. How is the critical value \( z_{\alpha/2} \) determined for different confidence levels?
2. How does increasing the sample size affect the margin of error?
3. Why does the confidence interval widen when the confidence level increases?
4. How would the results change if the sample proportion \( \hat{p} \) was closer to 0 or 1?
5. What happens to the margin of error if the population proportion is unknown?

**Tip:** The margin of error decreases as the sample size increases, because a larger sample provides more accurate estimates of the population proportion.

In the spring of 2017, the Consumer Reports National Research Center conducted a survey of 1,007 adults to learn about their major healthcare concerns. The survey results showed that 577 of the respondents lack confidence they will be able to afford health insurance in the future. (a) What is the point estimate of the population proportion of adults who lack confidence they will be able to afford health insurance in the future? (b) At 90% confidence, what is the margin of error? (c) Develop a 90% confidence interval for the population proportion of adults who lack confidence they will be able to afford health insurance in the future. (d) Develop a 95% confidence interval for this population proportion.

Learn how to calculate the population proportion of adults who lack confidence in affording health insurance, using data from a 2017 survey. This tutorial covers the point estimate, margin of error at 90% confidence, and the development of 90% and 95% confidence intervals.

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