The Age Distribution Of Householders Is An Important Tool

The Age Distribution Of Householders Is An Important Tool

The problem involves understanding the probability distribution of householders' ages, specifically focusing on the proportion of householders aged between 45 and 64 in 1995. A study estimates that 31% of all householders fall within this age range. A sample of 500 householders is taken, and the task is to employ the normal approximation to the binomial distribution to determine specific probabilities.

First, calculate the expected number of householders aged 45-64 in the sample. Since the estimated proportion (p) is 0.31, and the sample size (n) is 500, the expected number (mean) is:

μ = np = 500 × 0.31 = 155

The standard deviation (σ) of the binomial distribution is computed as:

σ = √(np(1-p)) = √(500 × 0.31 × 0.69) ≈ √(106.695) ≈ 10.33

Since n is large, apply the normal approximation with a continuity correction for the probabilities:

Part 1: Probability that fewer than 135 householders will be between 45 and 64

Using the continuity correction, we find P(X

Z = (X - μ) / σ = (134.5 - 155) / 10.33 ≈ -20.5 / 10.33 ≈ -1.984

Consulting standard normal distribution tables or using a calculator, P(Z

Part 2: Probability that between 135 and 180 householders (inclusive) will be between 45 and 64

We consider P(134.5

Z1 = (134.5 - 155) / 10.33 ≈ -1.984

Z2 = (180.5 - 155) / 10.33 ≈ 25.5 / 10.33 ≈ 2.471

Using the standard normal table, P(Z

P(135 ≤ X ≤ 180) ≈ P(Z

Hypothesis Test for Plant Heights

Kimmy has 34 plants with a normally distributed mature height pre-treatment with mean μ₀ = 14.8cm and standard deviation σ = 4.3cm. After applying a treatment, the sample mean height is 15.9cm, which is higher than the historical mean. The question is to determine if this observed difference is statistically significant at the 0.05 level.

Since the population standard deviation is known, perform a one-sample z-test:

Null hypothesis (H₀): μ = 14.8cm

Alternative hypothesis (H₁): μ > 14.8cm (indicating an increase due to treatment)

The test statistic (z) is calculated as:

z = (x̄ - μ₀) / (σ / √n) = (15.9 - 14.8) / (4.3 / √34) ≈ 1.1 / (4.3 / 5.831) ≈ 1.1 / 0.738 ≈ 1.49

Using standard normal tables, the p-value for this one-sided test with z = 1.49 is approximately 0.068 (since P(Z > 1.49) ≈ 0.068). This p-value exceeds the α level of 0.05, so we fail to reject the null hypothesis. There is insufficient evidence to conclude the treatment significantly increased plant height at the 5% significance level.

Estimating p-value from the t-statistic

Given a t-statistic of 2.3 and degrees of freedom n - 1 = 3, the goal is to find the p-value for a one-sided test where H₁: μ

Paper For Above instruction

The analysis of householders' age distribution provides valuable insights for marketers seeking to target age-specific demographics effectively. In 1995, a study estimated that 31% of householders were between 45 and 64 years old. Employing the normal approximation to the binomial distribution, with a sample size of 500, allows us to calculate the probabilities of observing certain counts within this age group. These probabilities inform about the likelihood of particular distributions occurring due to chance, which can be crucial in tailoring marketing strategies.

Calculating the expected number and standard deviation involves basic binomial properties. The expected number of householders aged 45-64, denoted as np, is 155, while the standard deviation, √(np(1-p)), is approximately 10.33. Using the continuity correction enhances the accuracy of probability estimates when approximating discrete distributions with continuous ones.

The probability that fewer than 135 householders are aged 45-64 is about 0.0235. This is computed by standardizing the value to a Z-score, which is approximately -1.984, and consulting the standard normal distribution table. Such a low probability indicates that observing 134 or fewer householders in this age group under the assumption that 31% is true is unlikely.

In contrast, the probability that between 135 and 180 householders (inclusive) belong to this age group is very high—about 0.9698. The Z-scores at 134.5 and 180.5 are approximately -1.984 and 2.471 respectively, translating to lower and upper bounds in the standard normal distribution. This high probability suggests that observing such counts is quite consistent with the estimated proportion.

In addition to demographic analysis, statistical inference plays a crucial role in agricultural studies. Kimmy's experiment on plant heights employs hypothesis testing to detect whether a treatment genuinely increases plant height. With a known population standard deviation, a z-test is appropriate. The observed mean height after treatment is 15.9cm, which is higher than the historical mean of 14.8cm, with a calculated z-value of about 1.49. The p-value exceeds 0.05, indicating that the increase is not statistically significant at the 5% level.

Furthermore, the estimation of p-values based on t-statistics provides a means to interpret statistical significance in smaller samples, particularly when population standard deviations are unknown. A t-statistic of 2.3 with 3 degrees of freedom corresponds to a p-value of approximately 0.0475 for a one-sided test. This underlines how the specific value of the test statistic influences the decision to accept or reject hypotheses, emphasizing the importance of precise calculations.

Overall, the combination of probabilistic modeling and hypothesis testing constitutes a powerful toolkit in statistics, enabling informed decisions in marketing, agriculture, and other fields. Understanding these concepts facilitates the interpretation of data and supports evidence-based strategies that can significantly impact business and scientific outcomes.

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