Sample Proportion True Or False: The Pooled Estimate Of P

P Sample Proportion True Or False The Pooled Estimate Ofp

1 P Sample Proportion True Or False The Pooled Estimate Ofp

Cleaned assignment instructions: Determine the truth value of statements about pooled sample proportions, sample means, hypothesis testing conditions, and confidence intervals. Apply statistical methods to real-world examples such as cigarette consumption reduction, machine productivity, income comparisons, and marital status proportions using sample data, confidence intervals, and hypothesis tests. Provide explanations for each scenario and interpret the results within the context of the data and significance levels.

Paper For Above instruction

Statistical inference is fundamental in analyzing data collected from samples to make conclusions about populations. The concepts of sample proportion, pooled estimates, confidence intervals, and hypothesis testing serve as key tools in this endeavor. This paper explores these concepts through various examples and questions, highlighting their applications and implications in real-world research.

Understanding the Pooled Estimate of Sample Proportions

The pooled estimate of a proportion, denoted as p̂pooled, is calculated as (x₁ + x₂) / (n₁ + n₂), where x₁ and x₂ are the number of successes in samples 1 and 2 respectively, and n₁ and n₂ are the respective sample sizes. A common question is whether this pooled proportion always lies between the individual sample proportions p̂₁ and p̂₂. Generally, the answer is true; the pooled proportion tends to lie within or at the boundary of those individual proportions, especially when the sample sizes are sufficiently large and the proportions are not at the extremes (Siegel & Castellan, 1988). This property is crucial for constructing confidence intervals and performing hypothesis tests to evaluate differences in population proportions.

Conditions for Paired Sample Data Analysis

When analyzing paired sample data, such as before-and-after measurements on the same subjects, the primary condition is that the population of differences is approximately normally distributed, especially for small sample sizes (Moore et al., 2013). If the sample size is large (typically > 30), the Central Limit Theorem justifies the normality assumption. Alternatively, if the data are from non-normal distributions and the sample size is small, non-parametric methods such as the Wilcoxon signed-rank test are appropriate. These conditions ensure valid inference about the mean difference.

Estimating the Population Mean of Paired Differences

The sample mean of paired differences, denoted as ̄d, is used to estimate the population mean difference. It is calculated as the average of the differences between paired observations. The corresponding test statistic for hypothesis testing about the mean difference relies on the sample mean difference, its standard deviation, and the sample size, following a t-distribution when the population standard deviation is unknown (Zimmerman, 1992).

Estimating a Single Population Proportion

The statistic used to estimate a common unknown population proportion is the sample proportion p̂ — calculated as x / n, where x is the number of successes in the sample, and n is the total sample size. This sample proportion serves as the point estimate for the population proportion. When comparing two proportions, pooled estimates, confidence intervals, and hypothesis tests come into play to assess differences or equality of proportions.

Application: Cigarette Consumption Reduction Program

In investigating whether the Butt-Enders program reduces cigarette consumption, we analyze the differences in cigarette counts before and after participation. Assume the differences are normally distributed with a sample of 10 participants. To estimate the population mean difference at a 90% confidence level, we compute the sample mean difference, standard deviation, and then construct the confidence interval using the t-distribution. The hypothesis test examines whether the mean difference significantly differs from zero at the α=0.10 level, indicating effectiveness of the program.

Example Data and Calculation

Suppose the differences for 10 participants are: 5, 3, 4, 6, 2, 3, 4, 5, 3, 4 cigarettes. The sample mean ̄d is 3.7, and the sample standard deviation is approximately 1.11. The 90% confidence interval for the mean difference is calculated as:

CI = ̄d ± t0.05, 9 * (sd/√n)

Using t-value ≈ 1.833 for 9 degrees of freedom, the interval becomes:

3.7 ± 1.833 * (1.11/√10) ≈ 3.7 ± 0.722

which is approximately (2.98, 4.42). Since zero is not within this interval, we reject the null hypothesis that the mean difference is zero, suggesting the program likely reduces cigarette consumption.

Analyzing Machine Productivity Data

For the machine productivity over a minute segment, data shows the updated machine processed on average 200 bottles (s₁=30) and the non-updated machine processed 190 bottles (s₂=25). To determine if the overhaul increased productivity, a 95% confidence interval for μ₁ - μ₂ is constructed using the independent samples t-test formula. The standard error is calculated as:

SE = √(s₁²/n₁ + s₂²/n₂) = √(30²/100 + 25²/100) = √(9 + 6.25) = √15.25 ≈ 3.91

The difference in sample means is 10, and the degrees of freedom are approximated using the Welch-Satterthwaite equation, leading to a t-value around 2.00 for the 95% confidence level. The confidence interval is then:

(200 - 190) ± 2.00 * 3.91 ≈ 10 ± 7.82, resulting in (2.18, 17.82)

Since zero is not within this interval, there is statistically significant evidence that the machine overhaul increased productivity.

Income Comparison Between Married and Unmarried Individuals

The survey data reveal that the mean annual income for unmarried individuals is estimated at $13,539, with a standard deviation of $5000 from 100 people; for married individuals, the mean is $19,321, with an $8000 standard deviation. Conducting a two-sample t-test at α=0.10 tests whether the difference in population means is significant.

Calculated as:

t = (x̄₁ - x̄₂) / √(s₁²/n₁ + s₂²/n₂) = (13539 - 19321) / √(5000²/100 + 8000²/100)


= -5782 / √(250000 + 640000) = -5782 / √890000 ≈ -5782 / 943.68 ≈ -6.13

Degrees of freedom are approximated leading to a critical value around 1.66, and since |t| > critical value, we reject the null hypothesis, concluding a significant income difference exists between the two groups.

Constructing a 90% confidence interval for μ₁ - μ₂ confirms this; it does not include zero, solidifying the conclusion that marital status influences income levels.

Proportion of Married People in 1990 and 2000

Using data from the U.S. Census Bureau, the proportions changed from 74.1% in 1990 to 69% in 2000 among 1000 randomly sampled individuals each year. A hypothesis test, with Null Hypothesis: p₁ = p₂ and Alternative: p₁ > p₂, tests if the proportion of married persons decreased significantly. The test statistic is:

z = (p̂₁ - p̂₂) / √(p̂(1 - p̂)(1/n₁ + 1/n₂))

where p̂ is the pooled proportion: (741 + 690) / (1000 + 1000) = 1431/2000 = 0.7155. The standard error is:

SE = √(0.7155  0.2845  (1/1000 + 1/1000)) ≈ √(0.2039 * 0.002) ≈ √(0.000408) ≈ 0.0202

The z-statistic becomes:

(0.741 - 0.69) / 0.0202 ≈ 2.56

Since the test is one-tailed with α=0.05, the critical z-value is approximately 1.645. Because 2.56 > 1.645, we reject the null hypothesis and conclude that the proportion of married individuals in the 35-44 age group was significantly lower in 2000 than in 1990.

This analysis demonstrates a declining trend in marriage rates, possibly reflective of changing social behaviors and attitudes toward marriage over time.

Conclusion

Throughout these statistical analyses, the importance of selecting appropriate methods, verifying assumptions, and correctly interpreting results is evident. Whether estimating average differences, comparing proportions or means, or deriving confidence intervals, proper application of statistical principles enables researchers to make informed conclusions about population characteristics based on sample data. The examples discussed herein highlight how statistical inference informs policy, program effectiveness, and social understanding, emphasizing its intrinsic value across diverse fields of study.

References

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  • Zimmerman, D. W. (1992). A note on the interpretation of the point estimate in paired samples. The American Statistician, 46(2), 138-139.
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