Math Problem Statement

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.

Question content area bottom

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.)

Solution

Part 1 (a) - Obtain a point estimate for the population proportion pp

The point estimate p^\hat{p} is the sample proportion of adults who believe that televisions are a luxury they could do without. This is calculated as:

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

Given:

  • Total adults surveyed n=10,121,012n = 10,121,012
  • Adults who believe televisions are a luxury x=513,513x = 513,513

p^=513,51310,121,0120.0507\hat{p} = \frac{513,513}{10,121,012} \approx 0.0507

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


Part 2 (b) - Verify the requirements for constructing a confidence interval about pp

The requirements for constructing a confidence interval about pp are:

  1. Random Sample: The sample can be assumed to be a simple random sample.

  2. Condition np^(1p^)10n\hat{p}(1 - \hat{p}) \geq 10: np^(1p^)=10,121,012×0.051×(10.051)n\hat{p}(1 - \hat{p}) = 10,121,012 \times 0.051 \times (1 - 0.051) np^(1p^)=10,121,012×0.051×0.949=489,476.643n\hat{p}(1 - \hat{p}) = 10,121,012 \times 0.051 \times 0.949 = 489,476.643

    Since 489,476.64310489,476.643 \geq 10, this condition is satisfied.

  3. Sample size relative to population: The sample size n=10,121,012n = 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:

    npopulation size=10,121,012300,000,0000.0337\frac{n}{\text{population size}} = \frac{10,121,012}{300,000,000} \approx 0.0337

    Since 0.03375%0.0337 \leq 5\%, this condition is also satisfied.

Therefore, the requirements for constructing a confidence interval about pp 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 np^(1p^)10n\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.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Statistics
Proportion Estimation
Confidence Intervals

Formulas

Sample Proportion (p̂) = x / n
n * p̂ * (1 - p̂) ≥ 10

Theorems

Central Limit Theorem

Suitable Grade Level

College Level (Introductory Statistics)