According to government data, 24 percent of teenagers in India under the age of 18 years live in households with incomes that are classified at a particular income level. A simple random sample of 400 teenagers in India under the age of 18 years was selected for a study of learning. If the government data is correct, which of the following best approximates the probability that at least 27 per cent of the teenagers in the sample live in households that are classified at a particular income level?
Understanding the Probability Question
The question asks us to find the probability that the proportion of teenagers in a simple random sample of 400 who live in households classified at a particular income level is at least 27%, given that the true population proportion is 24%. This is a classic problem involving the sampling distribution of a sample proportion.
Key Statistical Concepts
To solve this problem, we need to understand the following concepts:
- Population Proportion (p): The true proportion of individuals in the entire population who possess a certain characteristic. In this case, \(p = 0.24\) (24% of teenagers in India under 18 live in the specified income level households).
- Sample Proportion (ˆp): The proportion of individuals in a sample who possess the characteristic. In this case, we are interested in the probability that ˆp is at least 0.27.
- Sample Size (n): The number of individuals selected in the sample. Here, \(n = 400\).
- Sampling Distribution of the Sample Proportion: When the sample size is large enough (typically \(np \ge 10\) and \(n(1-p) \ge 10\)), the sampling distribution of ˆp can be approximated by a normal distribution.
- Mean of the Sampling Distribution: The mean of the sampling distribution of ˆp is equal to the population proportion, μ<sub>ˆp</sub> = p.
- Standard Error of the Sample Proportion: The standard deviation of the sampling distribution of ˆp is called the standard error, given by the formula SE(ˆp) = √(p(1-p)/n). It measures the typical distance of a sample proportion from the true population proportion.
- Z-score: A standardized score that tells us how many standard errors a particular value is away from the mean. For a sample proportion, the z-score is calculated as z = (ˆp - p) / SE(ˆp).
Checking Conditions for Normal Approximation
Before using the normal approximation, we check the conditions:
- \(np = 400 \times 0.24 = 96\)
- \(n(1-p) = 400 \times (1 - 0.24) = 400 \times 0.76 = 304\)
Since both \(np\) and \(n(1-p)\) are greater than or equal to 10, the normal approximation is appropriate.
Calculating the Z-score for the Sample Proportion
We are interested in the probability that the sample proportion ˆp is at least 0.27. We need to convert this value to a z-score using the formula:
\(z = \dfrac{\hat{p} - p}{\sqrt{\dfrac{p(1-p)}{n}}}\)
Substitute the given values:
- Population proportion \(p = 0.24\)
- Sample proportion of interest \(\hat{p} = 0.27\)
- Sample size \(n = 400\)
\(z = \dfrac{0.27 - 0.24}{\sqrt{\dfrac{0.24(1-0.24)}{400}}}\)
\(z = \dfrac{0.27 - 0.24}{\sqrt{\dfrac{0.24 \times 0.76}{400}}}\)
Finding the Probability
The question asks for the probability that at least 27% of teenagers in the sample live in the specified households. This is \(P(\hat{p} \ge 0.27)\). Using the z-score we calculated, this probability is equivalent to \(P(Z \ge z_{calculated})\). For a continuous distribution like the normal distribution, \(P(Z \ge z) = P(Z > z)\).
So, the probability is:
\(P\left(Z > \dfrac{0.27 - 0.24}{\sqrt{\dfrac{(0.24)(0.76)}{400}}}\right)\)
Comparing with Options
Let's compare our result with the given options:
| Option | Expression | Analysis |
|---|---|---|
| 1 | \(P\left(z>\dfrac{0.24-0.27}{\sqrt{\dfrac{(0.24)(0.76)}{400}}}\right)\) | Numerator is \((p - \hat{p})\) instead of \((\hat{p} - p)\). Incorrect. |
| 2 | \(P\left(z>\dfrac{0.27-0.24}{\sqrt{\dfrac{(0.50)(0.50)}{400}}}\right)\) | Uses 0.50 in the standard error formula instead of the population proportion \(p=0.24\). Incorrect. |
| 3 | \(P\left(z>\dfrac{0.27-0.24}{\sqrt{\dfrac{(0.27)(0.73)}{400}}}\right)\) | Uses the sample proportion \(\hat{p}=0.27\) in the standard error formula instead of the population proportion \(p=0.24\). Incorrect for calculating probability based on a known population proportion. |
| 4 | \(P\left(z>\dfrac{0.27-0.24}{\sqrt{\dfrac{(0.24)(0.76)}{400}}}\right)\) | Matches our calculated z-score expression and the correct probability direction \(P(Z > z)\). Correct. |
Option 4 correctly represents the probability calculation using the z-score for the sample proportion ˆp being greater than 0.27, given the population proportion \(p=0.24\) and sample size \(n=400\). The standard error is calculated using the population proportion \(p\).
Revision Table: Key Formulas
| Concept | Formula |
|---|---|
| Population Proportion | \(p\) |
| Sample Proportion | \(\hat{p} = \dfrac{\text{Number of successes}}{\text{Sample size}} = \dfrac{x}{n}\) |
| Standard Error of Sample Proportion | \(SE(\hat{p}) = \sqrt{\dfrac{p(1-p)}{n}}\) (used when population proportion p is known) |
| Z-score for Sample Proportion | \(z = \dfrac{\hat{p} - p}{SE(\hat{p})} = \dfrac{\hat{p} - p}{\sqrt{\dfrac{p(1-p)}{n}}}\) |
Additional Information: Normal Approximation Details
The normal approximation to the sampling distribution of the sample proportion ˆp is valid under certain conditions, typically \(np \ge 10\) and \(n(1-p) \ge 10\). These conditions ensure that the underlying binomial distribution (which models the number of successes in n trials) is sufficiently symmetric and bell-shaped to be approximated by a normal distribution. When these conditions are met, we can treat ˆp as a random variable following a normal distribution with mean μ<sub>ˆp</sub> = p and standard deviation σ<sub>ˆp</sub> = √(p(1-p)/n). This allows us to use z-scores and the standard normal table (or calculator) to find probabilities related to sample proportions.
It is important to use the population proportion \(p\) (0.24 in this problem) in the standard error formula when calculating probabilities based on a given population proportion. Using the sample proportion ˆp (0.27 in this problem) in the standard error formula is generally done when constructing confidence intervals, where the population proportion is unknown and estimated by the sample proportion.
Important Questions from Elementary Statistics
Demand for seats in a university is at its highest in the fall; demand also trends to grow and fall off in 25 year waves. In time service forecasting, the former demand characteristic would be called ______ and the latter would be called _______.
The system of combining two or more overlapping series of index numbers to obtain a single continuous series is called
The rise in the number of patients due to heatstroke is an example of:
Which index satisfies the factor reversal test?
Calculate the coefficient of range for the following series:
Item
10
12
14
16
18
20
22
Frequency
5
3
8
12
34
63
8
