All Calculations Should Be Done In Word Using Equatio 201195
All Calculations Should Be Done In Word Using Equation Editor And All
The following assignment involves analyzing multiple statistical scenarios using hypothesis testing, chi-square tests, Mann-Whitney (Wilcoxon) tests, and t-tests. The goal is to interpret data, formulate hypotheses, perform calculations accurately using Word's Equation Editor, and interpret the results to draw valid conclusions at specified significance levels. This comprehensive exercise emphasizes understanding statistical methods, executing proper calculations, and reporting findings clearly and accurately, utilizing both Word and Excel as needed.
Paper For Above instruction
The assignment comprises four primary scenarios, each involving different statistical tests to evaluate relationships and differences in datasets. The approaches include chi-square tests for independence, Mann-Whitney U tests for median comparisons, and t-tests for mean comparisons, all within a structured research context.
Scenario 1: Relationship Between Financial Condition and Education Level
The first scenario examines whether there is an association between adults' financial conditions and their education levels based on survey data conducted after the 1996 presidential election. The data is summarized in a contingency table and the task is to perform a chi-square test for independence.
Hypotheses:
Null hypothesis (H₀): Financial condition and education level are independent – there is no association.
Alternative hypothesis (H₁): Financial condition and education level are associated – there is a relationship.
Expected observations are calculated assuming independence by using the marginal totals:
Expected frequency for each cell = (Row total × Column total) / Grand total.
For example, if the total number of individuals with a High School degree is 300 and the total who are worse off now is 200 out of 800 respondents, then the expected frequency for the cell of "H.S. Degree – Worse off now" is (300 × 200) / 800 = 75.
The chi-square statistic is calculated as:
\[
\chi^2 = \sum \frac{(O - E)^2}{E}
\]
where O is the observed frequency, and E is the expected frequency for each cell.
Suppose the observed frequencies are given as follows:
Worse off now: H.S. Degree = 79, Lower Level = 111, Some College = 238, College Degree or Higher = 372.
No difference: H.S. Degree = 111, Lower Level = X, etc. (values to be computed).
Better off now: corresponding values.
After computing expected frequencies based on the marginal totals, plugging into the chi-square formula yields a test statistic which is then compared to the critical value from the chi-square distribution with appropriate degrees of freedom ((rows - 1) × (columns - 1)).
At a significance level of 0.05, if the calculated chi-square exceeds the critical value, we reject H₀, indicating a significant association.
Scenario 2: Comparing Median Waiting Times Between Two Banks
The second scenario involves testing whether there is a statistically significant difference in median waiting times during lunch hours between two branches, using the Mann-Whitney U test (Wilcoxon rank-sum test).
Hypotheses:
H₀: The median waiting times in the two branches are equal.
H₁: The medians are different.
Given samples are:
Bank 1: 4.21, 5.55, 3.02, 5.13, 4.77, 2.34, 3.54, 3.20, 4.50 (n₁=9)
Bank 2: 9.66, 5.90, 8.02, 5.79, 8.73, 3.82, 8.01, 8.35, 10.49, 6.68 (n₂=10)
The Wilcoxon U statistic is calculated by ranking all combined data, summing the ranks for one sample, and using the formula:
\[
U = R - \frac{n(n+1)}{2}
\]
where R is the sum of ranks for the sample. The critical value for U at α=0.05 is obtained from U tables for combined samples of these sizes.
If the calculated U is less than or equal to the critical U, we reject H₀, concluding that there is a significant difference in median waiting times.
Scenario 3: Testing Mean Income in a Cyclical Industry for a Company
The third scenario tests whether Wilco's recent net income data supports the claim that recent practices have improved performance. Using a one-sample t-test, the hypotheses are:
H₀: μ = 24 million (the prior average).
H₁: μ > 24 million (performance has improved).
Sample data: 30, 30, 30, 30, 30, 30 (n=6, x̄=30 million, s=10 million).
The t-statistic is calculated by:
\[
t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}
\]
with degrees of freedom df = n - 1 = 5.
The critical t-value at α=0.05 for a one-tailed test is obtained from t-distribution tables. If t exceeds this value, H₀ is rejected.
Summary
In all scenarios, proper execution involves calculating observed and expected frequencies, ranks, or means accurately using Word's Equation Editor. Critical values are sourced from standard statistical tables. The decision rule is straightforward: if the test statistic exceeds the critical value, reject the null hypothesis, suggesting statistically significant results. Accurate calculations and precise interpretation are vital in confirming or rejecting the hypotheses, ultimately contributing to more informed conclusions regarding economic relationships, business process improvements, or performance assessments in cyclical industries.
References
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