Data Gathered By The US Department Of Transportation

Data Gathered By The Us Department Of Transportation Show The Number

Analyze data related to travel distances in major metropolitan areas and statistical comparisons of test scores and household labor division, including hypothesis testing, confidence intervals, variance analysis, and proportion comparisons.

Paper For Above instruction

The provided problem set encompasses several statistical analyses pertinent to real-world data involving transportation usage, educational assessments, and household behavior. Each problem addresses different statistical concepts including confidence intervals, hypothesis testing for variances and proportions, and interpretation of statistical significance.

Firstly, addressing the transportation data, we examine the difference in average miles traveled per day by residents of Buffalo and Boston. A sample of 50 Buffalo residents reports a mean of 22.5 miles with a standard deviation of 8.4, while a sample of 100 Boston residents reports a mean of 18.6 miles with a standard deviation of 7.4. The point estimate for the difference in population means is simply the difference between sample means, which is 22.5 - 18.6 = 3.9 miles. This estimate suggests that Buffalo residents travel approximately 3.9 miles more daily than Boston residents on average.

Next, the construction of a 95% confidence interval for this difference involves calculating the standard error (SE) of the difference, given by:

SE = √(s1² / n1 + s2² / n2) = √(8.4² / 50 + 7.4² / 100) ≈ √(70.56 / 50 + 54.76 / 100) ≈ √(1.4112 + 0.5476) ≈ √1.9588 ≈ 1.399.

The critical t-value for a 95% confidence interval with degrees of freedom approximated (using conservative methods) is approximately 2.009 (for df ≈ 148), so the margin of error (ME) is:

ME = t* × SE ≈ 2.009 × 1.399 ≈ 2.810.

Thus, the 95% confidence interval for the difference in means is:

(3.9 - 2.810, 3.9 + 2.810) ≈ (1.09, 6.71) miles.

Because this interval does not contain zero, it suggests a statistically significant difference in daily travel distance, with Buffalo residents traveling more on average.

Regarding the assumptions and requirements for constructing this confidence interval, both samples are independent, randomly selected, and the sample sizes are sufficiently large to justify the use of the Central Limit Theorem, making the normal approximation appropriate. The standard deviations are known from the samples, and the data are approximately normally distributed, satisfying the conditions for inference.

Secondly, analyzing the variance comparison between male and female students' SAT scores, we conduct an F-test at the 1% significance level. The sample variances are:

Variance of females: 83² = 6889, based on n=121.

Variance of males: 69² = 4761, based on n=121.

Calculate the F-statistic as:

F = s1² / s2² = 6889 / 4761 ≈ 1.447.

The critical F-value at α=0.01 with (120,120) degrees of freedom is approximately 2.60 (from F-distribution tables). Since 1.447

The p-value for this F-test can be estimated using F-distribution tables or software, approximately p ≈ 0.2, which is greater than 0.01, confirming the non-rejection of the null hypothesis. If testing at a 5% significance level, the conclusion remains the same, as the p-value exceeds 0.05, indicating no significant difference in variances.

From the p-value, the lowest significance level at which the null hypothesis would be rejected is roughly 20%, since at α ≈ 0.2, the p-value threshold for rejection is met. This suggests the data supports the conclusion that variances are statistically similar at conventional significance levels.

Finally, regarding the comparison of perceived fairness in household division among men and women, the data involve proportions: 67.5% of men and 60.8% of women feel the division is fair. The hypotheses are:

H0: p_men ≤ p_women, versus HA: p_men > p_women.

Using a two-proportion z-test at the 5% level, the pooled proportion is:

p̂ = (0.675 × 1545 + 0.608 × 1691) / (1545 + 1691) ≈ (1044.375 + 1027.728) / 3236 ≈ 2072.103 / 3236 ≈ 0.640.

The standard error (SE) of the difference is:

SE = √(p̂(1 - p̂) (1/n1 + 1/n2)) ≈ √(0.64 × 0.36 (1/1545 + 1/1691)) ≈ √(0.2304 (0.000647 + 0.000591)) ≈ √(0.2304 × 0.001238) ≈ √0.000285 ≈ 0.0169.

The z-statistic is:

z = (0.675 - 0.608) / SE ≈ 0.067 / 0.0169 ≈ 3.96.

Corresponding to a p-value near 0.00003, far below 0.05, we reject H0 and conclude that a significantly higher proportion of men perceive the division as fair compared to women at the 5% significance level.

Using the confidence interval method, the 95% confidence interval for difference is:

Difference = 0.067, SE ≈ 0.0169, margin of error (MOE) = 1.96 × 0.0169 ≈ 0.033.

CI: (0.067 - 0.033, 0.067 + 0.033) = (0.034, 0.100). Since this interval lies entirely above zero, it confirms the test result that more men than women feel the division of housework is fair, with confidence.

In summary, the analyses collectively reinforce the conclusion that Buffalo residents travel significantly more miles daily than Boston residents, variances in SAT scores between genders are statistically similar, and more men perceive the housework division as fair than women, with high confidence and statistical significance.

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