Testing The Difference Between Two Means Using The Z-Test

Testing the Difference Between Two Means: Using the z-Test

Testing the difference between two means using the z-test involves comparing the sample means from two independent populations to determine if there is statistically significant evidence to suggest a true difference exists in their population means. This method is applicable when the sampling distributions are approximately normal, and population variances are known or sample sizes are sufficiently large for the Central Limit Theorem to hold. It is a common approach in various fields such as economics, psychology, and social sciences, often used to compare prices, test scores, or other quantitative measures across different groups or locations.

In this context, two scenarios are examined: one involving home prices in two different locations, and the other assessing whether local ACT scores differ significantly from the national average. The hypotheses in the first case test whether the average home prices are statistically equal versus the alternative that they differ. In the second case, the hypotheses evaluate whether the local ACT scores are significantly below the national average, indicating a potential performance discrepancy.

Scenario 1: Comparing Home Prices in Two Locations

The first scenario involves comparing the average home prices between Location A and Location B. The data provided include:

• Location A: mean = $93,430, standard deviation = $5,602, sample size = 35
• Location B: mean = $98,043, standard deviation = $4,731, sample size = 35

The primary question is whether the difference in average home prices is statistically significant at a significance level of α = 0.01. The null hypothesis (H₀) posits that the population means are equal:

H₀: μ_A = μ_B

While the alternative hypothesis (H₁) suggests they are not equal:

H₁: μ_A ≠ μ_B

To test this hypothesis, the z-test for independent samples is applied, using the formula:

z = (x̄_A - x̄_B) / √(σ_A²/n_A + σ_B²/n_B)

Substituting the values:

z = ($93,430 - $98,043) / √(($5,602)^2/35 + ($4,731)^2/35)

Calculating the numerator:

- $4,613

Calculating the denominator:

√( (31,389,604)/35 + (22,377,361)/35 ) = √(897,549 + 639,362) ≈ √1,536,911 ≈ 1,239.28

Then:

z ≈ -4,613 / 1,239.28 ≈ -3.72

Referring to the standard normal distribution table, with α = 0.01 for a two-tailed test, the critical z-values are approximately ±2.576. Since |z| = 3.72 > 2.576, we reject the null hypothesis and conclude that there is sufficient evidence to suggest a significant difference in home prices between the two locations at the 1% significance level.

Scenario 2: Evaluating ACT Scores

The second scenario assesses whether the local ACT scores are below the national average at a significance level of α = 0.05. The data are:

• National ACT score: mean = 21.4, standard deviation = 3.0, sample size = 1,000
• Local ACT score: mean = 20.8, standard deviation = 3.0, sample size = 500

The hypotheses are formulated as:

H₀: μ_local ≥ 21.4

H₁: μ_local

This is a one-tailed test, focusing on whether the local scores are significantly below the national average.

Applying the z-test for two independent samples, the z-statistic is calculated as:

z = (x̄_local - μ_national) / √(σ_local^2/n_local + σ_national^2/n_national)

Substituting the values:

z = (20.8 - 21.4) / √( (3.0)^2 / 500 + (3.0)^2 / 1000 ) = (-0.6) / √(9/500 + 9/1000)

Calculating the denominator:

√(0.018 + 0.009) = √(0.027) ≈ 0.164

Thus:

z ≈ -0.6 / 0.164 ≈ -3.66

For a one-tailed test at α = 0.05, the critical z-value is approximately -1.645. Since -3.66

Implications and Conclusion

The findings from both scenarios demonstrate the utility of the z-test in analyzing differences between two independent means. In the case of home prices, the significant difference suggests discrepancies in market values that may be attributed to factors like location-specific economic conditions or housing demand. For the ACT scores, the evidence indicates that students in the local area perform below the national average, which could prompt further investigations into educational resources or teaching methods.

It is important to note that the validity of these tests depends on assumptions such as the normality of the sampling distributions and known population variances. In real-world applications, if these conditions are not met, alternative tests like the t-test may be more appropriate. Nonetheless, the z-test remains a fundamental tool in statistical inference for comparing means when conditions are satisfied.

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