Determine Whether The Samples Are Independent Or Dependent

Determine whether the samples are independent or dependent

Identify whether the samples in each scenario are independent or dependent, and explain the reasoning behind your classification. Independent samples are those collected from separate, unrelated groups, while dependent samples are those where measurements are paired or related, such as before-and-after measurements on the same subjects.

1. The effectiveness of a new headache medicine is tested by measuring the amount of time before the headache is cured for patients who use the medicine and another group of patients who use a placebo drug.

Solution: This scenario involves two separate groups of patients—one receiving the medicine and the other receiving a placebo. Since the groups consist of different individuals and the measurements are collected independently, the samples are independent.

2. The effectiveness of a headache medicine is tested by measuring the intensity of a headache in patients before and after drug treatment.

Solution: Here, each patient serves as their own control, with measurements taken before and after treatment. These measurements are paired within each individual, so the samples are dependent.

Test for Difference in Proportions of Smokers in Two Age Groups

A study examines smoking habits in two age groups. In a random sample of 500 people aged 20-24, 22% are smokers, while in a sample of 450 people aged 25-29, 14% are smokers. The null hypothesis that the proportion of smokers is the same in both groups is tested at a significance level of 0.01. This involves a two-proportion z-test.

Step 1: State the hypotheses:

• Null hypothesis H₀: p₁ = p₂ (proportions are equal)
• Alternative hypothesis H_A: p₁ ≠ p₂ (proportions are not equal)

Step 2: Calculate sample proportions:

• p̂₁ = 0.22
• p̂₂ = 0.14

Step 3: Calculate the pooled proportion:

Pooled p̂ = (x₁ + x₂) / (n₁ + n₂) = (0.22×500 + 0.14×450) / (500 + 450) = (110 + 63) / 950 = 173 / 950 ≈ 0.1821

Step 4: Calculate the standard error (SE):

SE = √[p̂(1 - p̂) (1/n₁ + 1/n₂)]

SE = √[0.1821 × (1 - 0.1821) (1/500 + 1/450)] ≈ √[0.1821 × 0.8179 × (0.002 + 0.002222)] ≈ √[0.1821 × 0.8179 × 0.004222] ≈ √[0.000630] ≈ 0.0251

Step 5: Compute z-value:

z = (p̂₁ - p̂₂) / SE = (0.22 - 0.14) / 0.0251 ≈ 0.08 / 0.0251 ≈ 3.19

Step 6: Determine the critical value:

At α = 0.01 for a two-tailed test, critical z ≈ ±2.576.

Step 7: Make a decision:

Since |z| = 3.19 > 2.576, we reject the null hypothesis. There is sufficient evidence to conclude that the proportion of smokers differs between the two age groups.

Constructing a Confidence Interval for Difference of Means

A researcher assesses whether following a specific diet reduces blood pressure. The mean blood pressure in the treatment group (n₁=85) is 203.7 with a standard deviation of 38.7, and in the control group (n₂=203) is 39.2 with a standard deviation of 38.7. Using a 99% confidence level and the provided t-distribution table, the researcher aims to estimate the difference in mean blood pressures (μ₁ - μ₂).

Step 1: State the hypotheses that underpin the estimation rather than testing.

Null hypothesis: The true difference in means, μ₁ - μ₂, equals zero; the alternative is that it is not zero.

Step 2: Calculate the difference in sample means:

Difference = 203.7 - 39.2 = 164.5

Step 3: Compute standard error (SE) for the difference:

SE = √[(s₁² / n₁) + (s₂² / n₂)] = √[(38.7² / 85) + (39.2² / 203)]

SE ≈ √[(1497.69 / 85) + (1536.64 / 203)] ≈ √[17.62 + 7.58] ≈ √[25.20] ≈ 5.02

Step 4: Find critical t-value:

Degrees of freedom (approximate using Welch's formula):

df ≈ [(s₁² / n₁) + (s₂² / n₂)]² / [ ( (s₁² / n₁)² / (n₁-1) ) + ( (s₂² / n₂)² / (n₂-1) ) ]

Calculating df precisely yields a degree of freedom approximately around 125. Using t-tables, for 99% confidence and df ≈ 125, t_critical ≈ 2.62.

Step 5: Determine the margin of error:

ME = t_critical × SE = 2.62 × 5.02 ≈ 13.16

Step 6: Construct the confidence interval:

164.5 ± 13.16, which is (151.34, 177.66)

This interval suggests that, with 99% confidence, the true mean difference in blood pressure between the two groups is between approximately 151.34 and 177.66 mm Hg.

Testing the Effect of Tutoring on Math Scores

Five students' math scores before and after tutoring are compared to evaluate the tutoring's efficacy using a paired samples t-test at the 0.01 significance level. The data include:

• Student A: Before X_A1, After X_A2
• Student B: Before X_B1, After X_B2
• Student C: Before X_C1, After X_C2
• Student D: Before X_D1, After X_D2
• Student E: Before X_E1, After X_E2

Assuming the scores are provided, the following steps are performed:

Step 1: Hypotheses

• H₀: μ_d = 0 (no effect of tutoring)
• H_A: μ_d ≠ 0 (tutoring has an effect)

Step 2: Calculate the differences d_i = After - Before for each student; then compute the mean difference, \(\bar{d}\), and standard deviation s_d.

Step 3: Calculate the test statistic t = \(\bar{d}\) / (s_d / √n)

Step 4: Find the critical t-value for 4 degrees of freedom at 0.01 significance level (two-tailed): approximately ±4.604.

Step 5: Make a decision: if |t| > critical value, reject H₀.

Step 6: Interpret the results in non-technical terms, indicating whether tutoring significantly affected scores.