T-Tests and Confidence Intervals

Normal Sample Descriptive Statistics

• The 97.5% quantile of t distribution with 35 degrees of freedom is t0.975 = 2.030
• A t-interval procedure should be used to find the 95% Confidence Interval (CI) for the population mean µ because the standard deviation is not assumed known, and the sample standard deviation is used.
• For a 95% Confidence Interval, α=0.05, the critical value is t0.975 = 2.030 with df = 36 − 1 = 35
• The 95% Confidence Interval (CI) is (19.65, 22.35)
• Margin of Error M = 1.35

Hypothesis test for µ

• A one-sample (or one-mean) t-test should be used because the population standard deviation is not assumed known.
• The hypothesis test is two-tailed.
• The significance level is α = 5%.
• The value of the test statistic is t = 1.5.
• The critical values are ±t0.975 = ±2.030 with df = 35.
• The value of the test statistic does not fall in the rejection region, so the null hypothesis cannot be rejected.

Summary Statistics for Independent Random Samples from Two Populations

• The 97.5% quantile of t distribution with 28 degrees of freedom is t0.975 = 2.048.
• A two-tailed pooled t-test is performed with H0: μ₁ = μ₂ and Ha: μ₁ ≠ μ2 at 0.05 significance level.
• The population standard deviations are not assumed known and the sample standard deviations are very close.
• The test is two-tailed.
• The significance level is α = 5%.
• The value of the test statistic is t = -1.21 with Sp= 4.53.
• The critical values for two-tailed tests are ±t0.975 = ±2.048 with df = n₁ + n₂ − 2 = 28.
• The value of the test statistic –1.21 is inside the rejection region, so the null hypothesis cannot be rejected.
• The endpoints of the 95% Confidence Interval for μ₁ – μ₂ is −2 ± 3.39.
• The 95% Confidence Interval is (−5.39, 1.39).

One-Standard-Deviation x² Test for a Random Sample

• The sample standard deviation s = 6 and sample size n = 25.
• Two-tailed test with Ηο: σ = 5, Ηα: σ ≠ 5 at significance level α = 0.05
• The 5% and 97.5% quantiles of chi-square distribution with 24 degrees of freedom are X0.025 = 39.364 and X0.975 = 12.401, respectively.
• The critical values are X0.025 = 39.364 and X0.975 = 12.401.
• The value of the test statistic is x² = 34.56.
• The null hypothesis cannot be rejected because 12.401 2

Two-Standard-Deviations F Test for Two Independent Random Samples

• Given sample standard deviations and sample sizes: s₁ = 30.5, n₁ = 8, s₂ = 64.5, n₂ = 15.
• The 5% quantile of F distribution with df = (7, 14) is F0.05 = 2.76 and F0.95 = 0.283.
• Perform a left-tailed hypothesis test: H0: σ₁ = σ₂, Ha: σ₁ < σ₂ at significance level α = 0.05.
• The test is left-tailed.
• The value of the test statistic is F = 0.223.
• The critical value is F0.95 = 0.283 with df = (7, 14).
• The null hypothesis is rejected because 0.223 < 0.283.
• For a 90% Confidence Interval, the critical values are F0.05 and F0.95 with df = (7, 14) with values 2.76 and 0.283 respectively.
• The 90% Confidence Interval is (0.284, 0.888).

Estimating a Particular Proportion

• Given number of successes x = 28 and sample size n = 56.
• The 99.5% quantile of standard normal distribution is Z0.995 = 2.576.
• The sample proportion p= 0.5
• The 99% Confidence Interval is (0.328, 0.672)
• The margin of error for the Confidence Interval is E = 0.171.

Designing a Clinical Study

• Goal: Evaluate a therapy for head and neck cancer patients compared to the current standard therapy, based on the response rate (proportion of patients with significant tumor reduction or elimination).
• Current standard therapy response rate in the U.S. is 25%.
• The 95% quantile of standard normal distribution is Z0.95 = 1.645.
• Estimate the sample size with no prior knowledge of the response rate.
• Estimate the sample size needed with no prior knowledge.
• Margin of error for 90% Confidence Interval at most 10%.
• Critical values are ±1.645. -Need 68 head and neck cancer patients to achieve the margin of error 10%.
• For a sample of 75 patients there are 30 who showed significantly tumor reduction or complete elimination by the new drug
• The data is appropriate to use one-proportion z-interval procedure.
• The response rate from the sample is p=40%.
• The 90% Confidence Interval is (0.307, 0.493).
• The margin of error is Z0.95√(p(1 − p)/n) = 0.093 with a margin of error smaller than the required 10%.
• Hypothesis test is formulated with Ho: p = 0.25 and Ha: p > 0.25(right-tailed test).
• The significance level is α = 0.05 and the critical value is z0.95 = 1.645.
• The value of the test statistic is z=3.0
• Reject H0 to conclude that the data provide sufficient evidence that the new drug is better than the current standard of care in terms of response rate at 5% significance level, because 3.0 > 1.645, it falls in the rejection region for this right-tailed test.

Analyzing Data from a Randomized Study with Two Groups

• The objective is to compare two proportions given x₁ = 24, n₁ = 40, x₂ = 10, n₂ = 30.
• Right-tailed hypothesis test with α = 0.05, 90% Confidence Interval of p₁ – p₂.
• Z0.95 = 1.645.
• The sample proportions are p₁=0.60, p₂ = 0.33, pp= 0.486
• Use a two-proportions z-procedure because x₁ = 24, n₁ - x₁ = 16, x₂ = 10, n₂ - x₂ = 20, which are all is greater than 5.
• For the hypothesis test: Ho: p₁ = p₂ and Ha: p₁ > p₂
• The critical value is z0.95 = 1.645. with α = 0.05
• The value of the test statistic is z= 2.21
• The null hypothesis is rejected because 2.21 > 1.645.
• The 90% Confidence Interval of p1 -p2 is (0.08, 0.46) because α = 0.05.

One-Way ANOVA Analysis

• A statistician is performing a one-way ANOVA analysis, and only partial information is provided in the ANOVA table.
• The missing values of the provided table are completed
• 4 treatment groups (populations) are considered for the ANOVA analysis
• There are 24 observations included in all samples together.
• The null hypothesis is Ho: μ₁ = μ₂ = μ3 = μ4
• The data indicates there is not sufficient evidence to reject the null hypothesis because the F value is 2.5 < 3.1 with the critical value for the F-distribution is 3.1 at 5% level

Tukey Multiple Comparison

• Construction of a Tukey multiple comparison at the 95% family confidence level with simultaneous confidence intervals at 95% level.
• q0.95 = 4.16 with κ = 3 groups and v = 7 degrees of freedom.
• Recall the q-distribution parameters k = k and v = N - k, where k is the number of treatment groups and N is the total number of sample size.
• Given descriptive statistics for each sample like mean and sample size calculate descriptive statistics for all pair-wise population mean differences μ₁ – μ₂, μ₁ – μ₃, and μ₂ - μ3.
• SSE = 39.67
• MSE = 5.67
• The pairwise 95% simultaneous Confidence Intervals for μį – μj can be calculated as follows.
• μ₁ - μ₂: (-7.06, 4.39)
• μ₁ – μ₃: (-10.85, -0.15)
• μ₂ - μ₃: (-9.52, 1.19)