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Teens鈥 online profiles Over half of all American teens (ages 12 to 17 years) have an online profile, mainly on Facebook. A random sample of 487 teens with profiles found that 385 included photos of themselves.13 (a) Construct and interpret a 95% confidence interval for p. Follow the four- step process. (b) Is it plausible that the true proportion of American teens with profiles who have posted photos of themselves is 0.75? Use your result from part (a) to support your answer.

Short Answer

Expert verified
The 95% confidence interval for \( p \) is \((0.754, 0.826)\). A proportion of 0.75 is plausible.

Step by step solution

01

Define the Problem

We are tasked with determining a 95% confidence interval for the proportion \( p \) of all American teens with online profiles who include photos of themselves in their profiles. We have a sample size of \( n = 487 \) and a sample proportion \( \hat{p} = \frac{385}{487} \).
02

Check Conditions

To construct a confidence interval, we need to check certain conditions: 1. **Randomness**: The sample is a random sample of 487 teens.2. **Normality**: The sample size should be large enough so that both \( n \hat{p} \) and \( n(1 - \hat{p}) \) are at least 10. Calculate:\( n \hat{p} = 487 \times \frac{385}{487} = 385 \) \( n(1 - \hat{p}) = 487 - 385 = 102 \).Both values are greater than 10, so the normality condition is satisfied.
03

Calculate the Confidence Interval

The formula for the confidence interval for a proportion is:\[\hat{p} \pm z^* \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\]Where \( z^* \) is the z-score corresponding to a 95% confidence level (approximately 1.96). Substituting the known values, we get:\( \hat{p} = \frac{385}{487} \approx 0.7903 \).The standard error \( SE = \sqrt{\frac{0.7903 \times (1 - 0.7903)}{487}} \approx 0.0185 \).Thus, the 95% confidence interval is:\[0.7903 \pm 1.96 \times 0.0185\]This simplifies to \(0.7903 \pm 0.0363\), which is approximately \((0.754, 0.826)\).
04

Interpret the Confidence Interval

We are 95% confident that the true proportion of all American teens with online profiles who include photos of themselves is between 75.4% and 82.6%.
05

Evaluate Plausibility of 0.75

To determine if a true proportion of 0.75 is plausible, check if this value is within our confidence interval from part (a). Since 0.75 is within the interval \((0.754, 0.826)\), it is plausible that the true proportion could be 0.75.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Confidence Interval
In statistics, a confidence interval gives an estimated range of values which is likely to include an unknown population parameter. This range provides a level of confidence, such as 95%, about the estimate's precision.
To calculate a confidence interval for a proportion, we use the sample data to estimate the range within which the true population proportion lies. This requires:
  • The sample proportion (\( \hat{p} = \frac{385}{487} \approx 0.7903 \)).
  • The sample size (\( n = 487 \)).
  • A critical value from the standard normal distribution, usually denoted by \( z^* \), which is 1.96 for a 95% confidence interval.
The formula for the confidence interval is:\[\hat{p} \pm z^* \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\]By entering the sample proportion and sample size into this formula, we can create a range, making us 95% confident that the actual proportion of all American teens with online profiles, who include photos of themselves, falls within this span.
Confidence intervals help us understand the precision of our sample statistics and give valuable context to what the sampled data tells us about the larger population.
Proportion
A proportion in statistics is a type of ratio that compares a part to the whole. In this context, it helps us understand the part of American teens who have online profiles and post photos of themselves.
The proportion is calculated by dividing the number of teens who posted photos (385) by the total number of teens sampled (487), resulting in a sample proportion of \( \hat{p} \approx 0.7903 \).
Using this sample proportion, we can make estimates about the population proportion, which is the proportion of all American teens meeting this criterion. Proportions are essential for constructing confidence intervals and underlining the basis for further statistical testing to infer about the larger population beyond the sample.
Random Sampling
Random sampling is a method used to select a subset of individuals from a larger population. It is crucial to ensure that every individual has an equal chance of being chosen. This technique helps to produce representative data and reduces bias.
In the given exercise, a random sample of 487 teens was selected to estimate the proportion of teens including photos on their profiles. Random sampling provides credibility to the findings and allows statisticians to generalize from the sample to the larger population with greater accuracy.
  • Ensures that the sample reflects the diversity of the population.
  • Improves the reliability of statistical inference.
Random sampling is foundational to the entire process of making inferences about populations based on samples, as it improves the validity of conclusions drawn from statistical analysis.
Normality Condition
The normality condition is essential in constructing confidence intervals, particularly for a population proportion. It assures that the sampling distribution of the proportion is approximate to a normal distribution.For the sample size and proportion to satisfy this condition, both \( n \hat{p} \) and \( n(1 - \hat{p}) \) should be greater than or equal to 10. In our example, we calculate and find:
  • \( n \hat{p} = 487 \times 0.7903 = 385 \)
  • \( n(1 - \hat{p}) = 487 \times (1 - 0.7903) = 102 \)
Both these values exceed 10, thus ensuring the normality condition is satisfied. This is important because it allows the use of the normal distribution to create a confidence interval around the sample proportion.
When the normality condition is met, it validates treating the sampling distribution of the sample proportion as normal, facilitating more accurate and reliable interval estimation.

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