Conduct A Test At The Alpha Level Of Significance

Conduct A Test At The Alphaαequals010010level Of Significance By De

The assignment involves performing various hypothesis tests at a significance level of α = 0.10, including tests for differences between two proportions, constructing confidence intervals, and testing mean differences in matched-pairs data. The tasks include defining hypotheses, calculating test statistics, determining p-values, and interpreting results within the context of large sample data. Details include comparing proportions from independent samples, setting up appropriate null and alternative hypotheses, and understanding the criteria for statistical significance in different scenarios involving clinical and population data.

Paper For Above instruction

Hypothesis testing is a fundamental aspect of inferential statistics, enabling researchers to make decisions about populations based on sample data. In this paper, we address several hypothesis tests conducted at the 0.10 level of significance, focusing on two-proportion tests, confidence interval construction, and matched-pairs mean differences. Each scenario is examined with explicit formulation of hypotheses, calculation of test statistics, interpretation of p-values, and the implications of results.

Testing the Difference Between Two Proportions

The first scenario involves testing whether the proportion of successes in population 1 is greater than that in population 2 (i.e., p₁ > p₂). From the provided sample data, sample 1 has x₁=120 successes out of n₁=253 observations, and sample 2 has x₂=132 successes out of n₂=319 observations. The hypotheses are formulated as:

  • Null hypothesis (H₀): p₁ = p₂
  • Alternative hypothesis (H₁): p₁ > p₂

This represents a one-tailed test to determine if the first proportion exceeds the second.

To calculate the test statistic, we first estimate the pooled proportion:

p̂ = (x₁ + x₂) / (n₁ + n₂) = (120 + 132) / (253 + 319) = 252 / 572 ≈ 0.4406

Next, the standard error (SE) for the difference in proportions is computed as:

SE = √[p̂(1 - p̂)(1/n₁ + 1/n₂)] ≈ √[0.4406×0.5594(1/253 + 1/319)] ≈ √[0.2464(0.00395 + 0.00313)] ≈ √[0.2464×0.00708] ≈ √0.001745 ≈ 0.0418

The test statistic z is then:

z = (p₁̂ - p₂̂) / SE

where p₁̂ = x₁ / n₁ = 120 / 253 ≈ 0.4745, and p₂̂ = x₂ / n₂ ≈ 0.4138

So, z ≈ (0.4745 - 0.4138) / 0.0418 ≈ 0.0607 / 0.0418 ≈ 1.452

Consulting standard normal distribution tables, the p-value associated with z=1.452 (for a right-tailed test) is approximately 0.073. Since this p-value is less than the significance level 0.10, we reject H₀, providing evidence that p₁ > p₂.

Constructing a Confidence Interval for p₁ – p₂

Using the sample data with x₁=384, n₁=524 and x₂=414, n₂=558, a 90% confidence interval for the difference in proportions p₁ – p₂ is constructed as follows:

Calculate sample proportions:

p̂₁ = 384 / 524 ≈ 0.7333; p̂₂ = 414 / 558 ≈ 0.7410

The standard error (SE) for the difference is:

SE = √[p̂₁(1 – p̂₁)/n₁ + p̂₂(1 – p̂₂)/n₂] ≈ √[0.7333×0.2667/524 + 0.7410×0.2590/558] ≈ √[0.000373 + 0.000342] ≈ √0.000715 ≈ 0.0267

The critical value for 90% confidence (z*) is approximately 1.645. The margin of error (ME) is:

ME = z* × SE ≈ 1.645 × 0.0267 ≈ 0.044

The confidence interval is then:

(p̂₁ – p̂₂) ± ME = (–0.0077) ± 0.044, which yields approximately (–0.051, 0.037).

This interval includes zero, indicating no significant difference between the two population proportions at the 90% confidence level.

Testing Disease Incidence in a Clinical Trial

In a clinical trial with two groups of children—experimental and control—sample data reveal that in the experimental group, 75 out of 4000 children contracted the disease, and in the control group, 106 out of 4000 contracted it. The hypotheses are:

  • H₀: p₁ ≥ p₂
  • H₁: p₁

Where p₁ and p₂ denote the true proportions of children contracting the disease in the experimental and control groups, respectively.

Calculations proceed with the sample proportions:

p̂₁ = 75 / 4000 = 0.01875; p̂₂ = 106 / 4000 = 0.0265

The pooled proportion:

p̂ = (75 + 106) / (4000 + 4000) = 181 / 8000 = 0.0226

Standard error (SE):

SE = √ [p̂(1 – p̂)(1/n₁ + 1/n₂)] ≈ √ [0.0226×0.9774×(1/4000 + 1/4000)] ≈ √ [0.0226×0.9774×0.0005] ≈ √ [0.00001107] ≈ 0.00333

The z-test statistic:

z = (p̂₁ – p̂₂) / SE ≈ (0.01875 – 0.0265) / 0.00333 ≈ –0.00775 / 0.00333 ≈ –2.327

The p-value for this one-tailed test (p₁

Testing Mean Differences in Matched-Pairs Data

The researcher hypothesizes that the mean difference between paired observations from two related samples is less than zero, i.e., μd

For paired data, the test involves calculating the sample mean difference (d̄) and the standard deviation of differences (sₙ), then computing the t-statistic:

t = (d̄ – 0) / (sₙ / √n)

where n is the number of pairs.

If the calculated t is significantly negative with respect to the t-distribution with n–1 degrees of freedom, the researcher has statistical evidence to support μd

Overall, the hypothesis tests conducted at the 0.10 significance level help determine differences in proportions, means, and paired observations, providing insights into population parameters based on sample data. Their proper application requires careful formulation of hypotheses, accurate calculations, and nuanced interpretation considering the context of each study.

Conclusion

Statistical hypothesis testing at the 0.10 significance level is a crucial tool in empirical research. The examples examined illustrate the methodologies for comparing proportions, estimating confidence intervals, and analyzing matched pairs. Proper understanding and application of these techniques enable researchers to make informed decisions, support or refute hypotheses, and draw meaningful conclusions from sample data, all within the framework of large-sample theory and significance testing.

References

  • Altman, D. G. (1994). Practical Statistics for Medical Research. CRC Press.
  • DeGroot, M. H., & Schervish, M. J. (2014). Probability and Statistics (4th ed.). Pearson.
  • Newcombe, R. G. (1998). Two-sided confidence intervals for the binomial proportion: comparison of eleven methods. Statistics in Medicine, 17(8), 873-890.
  • Portney, L. G., & Watkins, M. P. (2015). Foundations of Clinical Research: Applications to Practice (3rd ed.). F.A. Davis Company.
  • Siegel, S., & Castellan, N. J. (2006). Nonparametric Statistics for the Behavioral Sciences. McGraw-Hill.
  • Smith, M. (2020). Statistical Methods in Medical Research. Wiley.
  • Snedecor, G. W., & Cochran, W. G. (1989). Statistical Methods (8th ed.). Iowa State University Press.
  • Wilks, S. S. (2011). Mathematical Statistics. Academic Press.
  • Zar, J. H. (2010). Biostatistical Analysis (5th ed.). Pearson.
  • Agresti, A., & Franklin, C. (2009). Statistics: The Art and Science of Learning from Data. Pearson.

Related Assignments