T-Test For Two Samples: A Drug Company Is Measuring Levels ✓ Solved

t-Test for Two Samples A drug company is measuring levels of

t-Test for Two Samples A drug company is measuring levels of oxygenation in patients after receiving a test medication. As the researcher, you are interested in whether Group I, which received the medication, has the same oxygenation levels as Group II, which did not. 1. Group 1: 2,3,3,4,4,7,8,9 2. Group 2: 1,2,2,3,4,4,5,5,6,8 Use Excel to run a t-test for two samples, assuming equal variances, with an alpha value of 0.05. Run the t-test and note: 1. Are you doing a one-tailed test or a two-tailed test (Excel will give you both)? 2. What is the probability that Group I is different from Group 2, using the p value? Is it significant against the benchmark of p

Paper For Above Instructions

Introduction and research question. The scenario asks whether the mean oxygenation level in patients who received a test medication (Group 1) differs from the mean oxygenation level in patients who did not receive the medication (Group 2). A two-sample t-test with equal variances is an appropriate inferential procedure for comparing the means of two independent groups when the outcome is approximately normally distributed and the sample sizes are modest to reasonable (Field, 2013; Gravetter & Wallnau, 2017). The goal is to determine whether any observed difference in sample means is unlikely to have occurred by random sampling variation alone under the null hypothesis of equal population means (H0: μ1 = μ2) (Motulsky, 2014). This analysis presumes independent samples and similar variances across groups, although alternatives exist when those assumptions are violated (Montgomery & Runger, 2014).

Data description and descriptive statistics. The data provided are Group 1 (n1 = 8) with values 2, 3, 3, 4, 4, 7, 8, 9 and Group 2 (n2 = 10) with values 1, 2, 2, 3, 4, 4, 5, 5, 6, 8. Calculating means and dispersion measures yields: Group 1 mean = 5.0, Group 2 mean = 4.0. The sample variances are s1^2 ≈ 6.857 and s2^2 ≈ 4.444, corresponding to standard deviations s1 ≈ 2.62 and s2 ≈ 2.11 (Field, 2013; Zar, 2010).

Method: two-sample t-test with equal variances. Under the equal-variances (pooled) assumption, the pooled variance is Sp^2 = [ (n1 − 1)s1^2 + (n2 − 1)s2^2 ] / (n1 + n2 − 2) = [7×6.857 + 9×4.444] / 16 = 88 / 16 = 5.50, so Sp ≈ 2.35. The standard error of the difference in means is SE = sqrt( Sp^2 [1/n1 + 1/n2] ) = sqrt(5.50 × (1/8 + 1/10)) ≈ sqrt(1.2375) ≈ 1.11. The t-statistic is t = (M1 − M2)/SE = (5.0 − 4.0)/1.11 ≈ 0.90, with degrees of freedom df = n1 + n2 − 2 = 16 (Field, 2013; Gravetter & Wallnau, 2017; Snedecor & Cochran, 1989).

Results from the t-test. With df = 16, the two-tailed p-value associated with t ≈ 0.90 is about 0.38–0.40, indicating that the observed difference is not statistically significant at the α = 0.05 level. The one-tailed p-value would be approximately half of the two-tailed value, near 0.19–0.20, depending on the direction specified in the alternative hypothesis (Field, 2013; Gravetter & Wallnau, 2017). Excel’s Data Analysis Toolpak (Two-Sample Assuming Equal Variances) typically reports t-statistic, along with P(T≤t) values that can be interpreted as one- or two-tailed depending on the reported option, enabling researchers to confirm the same conclusion (Field, 2013; Zar, 2010).

Interpretation and conclusions. The calculated t-statistic of about 0.90 with 16 degrees of freedom yields a two-tailed p-value well above 0.05, so we fail to reject the null hypothesis that the population means are equal. In practical terms, the data provide no convincing evidence that the medication altered oxygenation levels relative to the control group, given the sample sizes and observed variability. This interpretation aligns with the conventional criterion that p-values exceeding 0.05 do not provide evidence against H0 in the two-sample context (Cohen, 1988; Field, 2013). It is important to note, however, that the statistical power of this test depends on the true effect size, sample sizes, and variance, and the current study may be underpowered to detect moderate differences (Altman, 1990; Gravetter & Wallnau, 2017).

Assumptions, limitations, and considerations for future research. The analysis presented relies on the assumption of equal variances between groups, independence of observations, and approximate normality of the outcome within each group. When variances are unequal, Welch’s t-test provides a robust alternative that does not pool variances (Montgomery & Runger, 2014). If future studies are conducted with larger samples or with greater precision in outcome measurement, the test’s power to detect clinically meaningful differences will improve. Reporting should include a check for homogeneity of variances (e.g., Levene’s test) and a discussion of whether deviations from normality are a concern given the sample sizes (Rosner, 2010; Pagano & Gauvreau, 2000).

Practical steps to replicate in Excel. To replicate the analysis, input Group 1 and Group 2 data into two columns and use Excel’s Data Analysis Toolpak, “t-Test: Two-Sample Assuming Equal Variances.” The output provides t-statistic, degrees of freedom, and the P-value for the two-tailed test. If one-tailed testing is specified in the alternative hypothesis, interpret P(T≤t) accordingly in light of the direction of the effect. Researchers should report M, SD for each group, the t-statistic with df, and the exact p-value in line with APA guidelines for statistical reporting (Field, 2013; Gravetter & Wallnau, 2017).

Statistical reporting and implications for practice. In reporting, one would present the results as: t(df = 16) = 0.90, p = .38 (two-tailed); Group 1 mean 5.0 (SD ≈ 2.62), Group 2 mean 4.0 (SD ≈ 2.11); N1 = 8, N2 = 10. This form aligns with common APA-style formulations and is consistent with standard statistical guidelines for reporting t-tests (Rosner, 2010; Gravetter & Wallnau, 2017). If the goal is to claim clinical significance rather than statistical significance, researchers might consider reporting confidence intervals for the mean difference and discussing whether observed differences meet a predefined clinical threshold (Cohen, 1988; Motulsky, 2014).

Conclusion. The current analysis, based on the data provided, does not show a statistically significant difference in oxygenation levels between patients who received the medication and those who did not, under the assumption of equal variances and the chosen alpha level. While this finding suggests no detectable effect in this sample, it should not be interpreted as definitive proof of no effect; rather, it highlights the need for larger, well-controlled studies to achieve adequate power to detect plausible effect sizes (Altman, 1990; Field, 2013). Researchers should also consider alternate statistical approaches if underlying assumptions are challenged or if multiple outcomes or repeated measures are involved (Montgomery & Runger, 2014).

References

  1. Field, A. (2013). Discovering Statistics Using IBM SPSS Statistics (4th ed.). SAGE.
  2. Motulsky, H. (2014). Intuitive Biostatistics (3rd ed.). Oxford University Press.
  3. Zar, J. H. (2010). Biostatistical Analysis (5th ed.). Pearson.
  4. Snedecor, G. W., & Cochran, W. G. (1989). Statistical Methods (8th ed.). Iowa State University Press.
  5. Montgomery, D. C., & Runger, G. C. (2014). Applied Statistics and Probability for Engineers (6th ed.). Wiley.
  6. Altman, D. G. (1990). Practical Statistics for Medical Research. Chapman & Hall/CRC.
  7. Cohen, J. (1988). Statistical Power Analysis for the Behavioral Sciences (2nd ed.). Lawrence Erlbaum Associates.
  8. Gravetter, F. J., & Wallnau, L. B. (2017). Statistics for the Behavioral Sciences (11th ed.). Cengage.
  9. Pagano, M., & Gauvreau, K. (2000). Principles of Biostatistics. Duxbury Press.
  10. Rosner, B. (2010). Fundamentals of Biostatistics (7th ed.). Cengage Learning.

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