A random sample of

10121012

adults in a certain large country was asked​ "Do you pretty much think televisions are a necessity or a luxury you could do​ without?" Of the

10121012

adults​ surveyed,

513513

indicated that televisions are a luxury they could do without. Complete parts​ (a) through​ (e) below.

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Part 1

​(a) Obtain a point estimate for the population proportion of adults in the country who believe that televisions are a luxury they could do without.

ModifyingAbove p with caretpequals=0.5070.507

​(Round to three decimal places as​ needed.)

Part 2

​(b) Verify that the requirements for constructing a confidence interval about p are satisfied.

The sample

can be assumed to be

a simple random​ sample, the value of

n ModifyingAbove p with caret left parenthesis 1 minus ModifyingAbove p with caret right parenthesisnp1−p

is

enter your response here​,

which is

greater than or equal to

​10, and the

sample size

can be assumed to be

less than or equal to​ 5% of the

population size.

​(Round to three decimal places as​ needed.)

### Part 1 (a) - Obtain a point estimate for the population proportion \( p \)
The point estimate \( \hat{p} \) is the sample proportion of adults who believe that televisions are a luxury they could do without. This is calculated as:

\[
\hat{p} = \frac{\text{Number of adults who think televisions are a luxury}}{\text{Total number of adults surveyed}}
\]

Given:
- Total adults surveyed \( n = 10,121,012 \)
- Adults who believe televisions are a luxury \( x = 513,513 \)

\[
\hat{p} = \frac{513,513}{10,121,012} \approx 0.0507
\]

So, the point estimate \( \hat{p} = 0.051 \) (rounded to three decimal places).

---

### Part 2 (b) - Verify the requirements for constructing a confidence interval about \( p \)

The requirements for constructing a confidence interval about \( p \) are:

1. **Random Sample**: The sample can be assumed to be a simple random sample.
2. **Condition \( n\hat{p}(1 - \hat{p}) \geq 10 \)**:
   \[
   n\hat{p}(1 - \hat{p}) = 10,121,012 \times 0.051 \times (1 - 0.051)
   \]
   \[
   n\hat{p}(1 - \hat{p}) = 10,121,012 \times 0.051 \times 0.949 = 489,476.643
   \]

   Since \( 489,476.643 \geq 10 \), this condition is satisfied.

3. **Sample size relative to population**: The sample size \( n = 10,121,012 \) should be less than or equal to 5% of the population size. If we assume the population size of the country is large (say, 300 million or more), then:

   \[
   \frac{n}{\text{population size}} = \frac{10,121,012}{300,000,000} \approx 0.0337
   \]

   Since \( 0.0337 \leq 5\% \), this condition is also satisfied.

Therefore, the requirements for constructing a confidence interval about \( p \) are satisfied.

---

Do you want to continue with part (c)? Let me know if you have any questions!

---

Here are five follow-up questions related to this problem:

1. How is the sample proportion different from the population proportion?
2. What is the significance of the condition \( n\hat{p}(1 - \hat{p}) \geq 10 \) in confidence interval estimation?
3. How would increasing the sample size affect the margin of error in this survey?
4. What statistical methods can be used to determine an appropriate sample size for surveys?
5. How does the sample size influence the precision of the confidence interval?

**Tip**: When calculating proportions, always ensure the sample is representative of the population to avoid bias.

A random sample of 10,121,012 adults in a certain large country was asked 'Do you pretty much think televisions are a necessity or a luxury you could do without?' Of the 10,121,012 adults surveyed, 513,513 indicated that televisions are a luxury they could do without. Complete parts (a) through (e) below. (a) Obtain a point estimate for the population proportion of adults in the country who believe that televisions are a luxury they could do without. (b) Verify that the requirements for constructing a confidence interval about p are satisfied.

Solve a real-world problem involving a survey of 10,121,012 adults, where 513,513 indicated televisions are a luxury. Learn how to estimate the population proportion and verify the requirements for constructing a confidence interval. Suitable for introductory statistics students.

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