Assignment 4: Hypothesis Testing On Means And Proportions

Assignment 4hypothesis Testing On Means And Proportions1 The Hourly

The assignment involves conducting hypothesis tests on various datasets related to wages, weights, incomes, stocking levels, germination capacities, variances, and distributions. Tasks include testing the mean wages against a known population mean, verifying average weights against a standard, comparing incomes of professionals, evaluating stocking levels in forestry plots, testing seed lot germination rates, estimating population variances, analyzing distribution fit, and assessing dependence between variables such as accidents and sex, or grades and year. The objective is to apply different statistical hypothesis testing techniques—such as z-tests, t-tests, chi-square tests, and variance tests—at specified significance levels to determine whether observed data support or refute certain claims or assumptions.

Paper For Above instruction

Introduction

Hypothesis testing is a fundamental statistical tool used to make inferences about population parameters based on sample data. It involves proposing a null hypothesis (H₀) representing a default or status quo assumption, and an alternative hypothesis (H₁ or Ha) that reflects a new claim or different state of the population. By analyzing sample data, researchers determine whether there is enough evidence to reject the null hypothesis at a pre-specified significance level (α). This paper discusses several hypothesis testing scenarios involving means, proportions, variances, and distributions, illustrating their application in forestry, industry wages, income comparisons, seed germination, and more.

Testing the Mean Hourly Wage

The first scenario examines whether a forestry company pays its summer students inferior wages. The population mean wage is known to be $13.20 with a standard deviation of $2.50, and a sample of 40 students earns an average of $12.20. The hypothesis test is set as follows:

- Null hypothesis (H₀): μ = $13.20

- Alternative hypothesis (H₁): μ

The test statistic for a z-test is calculated as:

\[ z = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}} \]

Calculating:

\[ z = \frac{12.20 - 13.20}{2.50 / \sqrt{40}} \approx \frac{-1.00}{0.395} \approx -2.53 \]

At α = 0.05, the critical z-value for a one-tailed test is approximately -1.645. Since -2.53

Testing the Drained Weight of Cans

A sample of 15 cans reveals drained weights to determine if they meet the 12-ounce standard. Using a 1% significance level, the null hypothesis is that the mean weight is 12 ounces versus the alternative that it is not 12 ounces. The sample data, standard deviation, and sample size enable conducting a t-test assuming the population standard deviation is unknown or the sample standard deviation is used. The test statistic is:

\[ t = \frac{\bar{x} - \mu_0}{s/ \sqrt{n}} \]

Based on the data, we would compute the sample mean and standard deviation, then compare the calculated t to the critical t-value at 14 degrees of freedom for α = 0.01. If the test statistic falls outside the acceptance region, we reject the null hypothesis, indicating a deviation from the 12-ounce standard.

Income Comparison between Engineers and Foresters

This scenario compares the average annual income after ten years for engineers and foresters, with sample data provided for two groups. The test aims to determine whether there's a significant difference in mean incomes:

- Null hypothesis: μ₁ = μ₂

- Alternative hypothesis: μ₁ ≠ μ₂

Using a two-sample t-test for unequal variances, the test statistic is:

\[ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \]

Calculating the t-value and degrees of freedom, then comparing to the critical t at α = 0.05, enables decision-making about the income disparity.

Assessment of Stocking Levels

In forestry, stocking exceeds the 80% threshold in a surveyed area with 85 plots examined. The null hypothesis states that the proportion of stocked plots is 80% versus the alternative that it exceeds 80%. Using a z-test for proportions:

\[ z = \frac{\hat{p} - p_0}{\sqrt{p_0(1 - p_0)/n}} \]

where \(\hat{p} = 85/110 ≈ 0.7727\). Since the proportion is actually less than 80%, the test at α = 0.01 would determine if the observed proportion significantly deviates from the threshold, leading to conclusions about the area's regeneration status.

Proportion Comparison of Stocking Levels in Different Areas

A second area shows 84% stocking across 100 plots. The question is whether this proportion differs significantly from the previous area. A hypothesis test for difference in proportions compares the two:

- Null hypothesis: proportions are equal (p₁ = p₂)

- Alternative hypothesis: proportions are not equal

Using a z-test for two proportions, the combined proportion and standard error are calculated, and the z-test determines if the difference is statistically significant at α = 0.05.

Germination Capacity of Seeds

Testing if a seed lot with 8 germinated seeds out of 18 meets the minimum acceptable rate of 70%. The null hypothesis: p ≥ 0.70; the alternative: p

\[ z = \frac{\hat{p} - p_0}{\sqrt{p_0(1 - p_0)/n}} \]

with \(\hat{p} = 8/18 \approx 0.444\), to determine if the seed lot should be accepted or rejected at α = 0.05.

Difference and Similarity Between Interval Estimation and Hypothesis Testing

Interval estimation and hypothesis testing are related but serve different purposes. Interval estimation constructs a range (confidence interval) within which the true population parameter likely falls, providing a measure of estimation precision. Conversely, hypothesis testing evaluates specific claims or assumptions about the population by determining whether sample data provide sufficient evidence to reject a null hypothesis. Both methods are based on similar statistical foundations—sampling distributions and significance levels—but interval estimation emphasizes estimation accuracy, while hypothesis testing focuses on hypothesis verification or falsification.

Estimating Variability in Forest Stand Volume Increment

Given volume increments from thinned and unthinned plots, the goal is to estimate the standard deviation of the thinned stand's volume increment with 90% confidence. Using sample data, the sample standard deviation is computed, and the chi-square distribution provides the confidence interval for the variance, whose square root yields the standard deviation.

Comparing Variances Between Forest Stands

To determine whether the variances from thinned and unthinned stands are significantly different, an F-test is employed:

\[ F = \frac{s_1^2}{s_2^2} \]

The critical F-value at 90% confidence (α = 0.10) with appropriate degrees of freedom tests homogeneity of variances. If the calculated F exceeds the critical value, variances are significantly different.

Distribution of Tossing Coins Until a Tail

This scenario models the number of tosses until a tail appears, which follows a geometric distribution:

- Distribution: Geometric distribution with parameter p, the probability of tail.

- Probability: \( P(X = k) = (1 - p)^{k-1} p \)

The hypothesis test at α = 0.05 assesses whether observed data fit this geometric distribution, typically through a goodness-of-fit test such as the chi-square test.

Normality of Lengths of Douglas-fir Cones

Using the frequency distribution of cone lengths, a chi-square goodness-of-fit test assesses if the data follow a normal distribution. The expected frequencies are computed based on a normal distribution fitted to the data, and the chi-square statistic compares observed and expected counts for each class, testing at α = 0.05.

Analysis of Fatal Accidents and Sex Dependence

Contingency table analysis examines if fatal accidents depend on sex. The chi-square test for independence compares observed and expected frequencies under the assumption of independence at α = 0.05. Significant differences suggest sex influences the likelihood of fatal accidents.

Assessing Grade Distributions for Independence Over Years

Finally, the examination of grade distributions over three years employs a chi-square test of independence. The null hypothesis states that grade distributions are independent of the year, and the test determines if observed variations are statistically significant at α = 0.01.

Conclusion

The diverse applications of hypothesis testing across these cases demonstrate its utility in forestry, industry, education, and safety analysis. Accurate interpretation of test results informs decision-making, policy formulation, and resource management, emphasizing the importance of selecting appropriate tests, understanding their assumptions, and analyzing the results within context.

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