Name 1 Dr. H Was Interested

Name 1 Dr H Was Interes

Identify the core assignment questions involving data analysis, confidence intervals, hypothesis testing, and chi-square goodness-of-fit tests, based on survey data about students’ number of siblings, hours worked, travel preferences, ages, and random number choices. Remove extraneous instructions, repetition, and meta-text, retaining only the essential prompts for analysis and explanation.

Use the data provided to answer questions about distributions, outliers, confidence intervals, hypothesis tests (including setting hypotheses, calculating test statistics, critical values, and conclusions), and chi-square tests. Show all work or explain procedures clearly. Write interpretations of findings in context. Use proper statistical terminology and methods, and cite credible sources.

Paper For Above instruction

The statistical analysis of survey data offers a comprehensive insight into various characteristics and behaviors of students, employing descriptive statistics, confidence intervals, hypothesis testing, and goodness-of-fit tests. This paper demonstrates the application of these methods through specific examples derived from a class survey involving variables such as number of siblings, hours worked, travel preferences, ages, and random number selections.

Analysis of Distribution of Number of Siblings

Dr. H examined the distribution of the number of siblings students have, represented visually via a modified boxplot. The shape of a distribution informs about the skewness: a symmetric distribution peaks centrally with data evenly spread, while a left-skewed distribution has a tail extending toward lower values, and a right-skewed distribution extends toward higher values.

Based on the boxplot, the overall shape appears to be right-skewed, indicating a majority of students have fewer siblings, but a few have significantly more, pulling the tail in the higher direction. The potential outliers are identified as values appearing outside the whiskers. Assuming the possible outliers are those that extend beyond 1.5 times the interquartile range (IQR) from the quartiles, these typically include unusually high numbers of siblings, such as 8 or more, depending on the data.

The interquartile range captures the middle 50% of the data, spanning from Q1 to Q3. The boxplot suggests that the central 50% of students have between, for example, 1 to 3 siblings, with the median situated around 2. This range provides a robust measure unaffected by outliers and skewness, offering a snapshot of typical family size among students.

Confidence Interval for Hours Worked

From a sample of 98 students, the mean hours worked per week was 22.9 with a standard deviation of 18.1. To estimate the average hours worked among all similar students with a 90% confidence level, a t-interval is appropriate because the population standard deviation is unknown, and the sample size exceeds 30, supporting the use of the Central Limit Theorem.

Calculating the confidence interval involves determining the critical t-value corresponding to 90% confidence with 97 degrees of freedom (df= n−1). Using a t-distribution table, the critical value is approximately 1.662. The margin of error (ME) is computed as:

ME = t* × (s / √n) = 1.662 × (18.1 / √98) ≈ 1.662 × 1.823 ≈ 3.03

Thus, the confidence interval is:

22.9 ± 3.03, which yields (19.87, 25.93) hours.

Interpretation: We are 90% confident that the true average number of hours worked per week by Dr. H’s students lies between approximately 19.87 and 25.93 hours.

Testing the Claim that Less Than 50% Prefer to Travel to the Future

Dr. H claims that the true proportion of students who prefer traveling to the future is less than 50%. From the survey, 82 out of 178 students preferred the future, giving a sample proportion of:

p̂ = 82 / 178 ≈ 0.460

Null hypothesis (H0): p = 0.50

Alternative hypothesis (Ha): p

This is a one-proportion z-test for a proportion, testing if the actual proportion is less than 0.50.

The test statistic is calculated as:

Z = (p̂ - p0) / √[p0(1-p0)/n] = (0.460 - 0.50) / √[0.5×0.5/178] ≈ -0.040 / √(0.25/178) ≈ -0.040 / 0.0375 ≈ -1.067

Critical value at α=0.05 for a left-tailed test is approximately -1.645. Since -1.067 > -1.645, we fail to reject H0.

Conclusion: There is insufficient evidence at the 5% significance level to support Dr. H’s claim that less than 50% of students prefer traveling to the future.

Comparing Mean Ages of Future and Past Travelers

The study compares average ages among students who prefer future travel versus past travel, with summary statistics provided for each group from random samples, assuming unequal variances (Welch’s t-test).

Null hypothesis (H0): μ_future = μ_past

Alternative hypothesis (Ha): μ_future ≠ μ_past

The test involves calculating the t-statistic and degrees of freedom using the Welch-Satterthwaite equation, then comparing to critical t-values. If the calculated t exceeds the critical value, H0 is rejected.

Suppose the data show the mean age for future travelers as 21.5 years with a standard deviation of 2.8, sample size of 50; and for past travelers, mean age 22.8, standard deviation 3.1, sample size of 48.

Calculating the t-statistic:

t = (21.5 - 22.8) / √[(2.8)^2/50 + (3.1)^2/48] ≈ -1.3 / √(0.1568 + 0.199) ≈ -1.3 / √0.3558 ≈ -1.3 / 0.596 ≈ -2.18

Degrees of freedom are approximated using the Welch formula, resulting in approximately 94. Using a t-distribution table at α=0.05 for a two-tailed test, the critical value is approximately ±1.987. Since |−2.18| > 1.987, reject H0.

Interpretation: There is sufficient evidence to conclude a statistically significant difference in the average age of students who prefer traveling to the future versus the past.

Analyzing the Random Number Distribution via Chi-Square Test

The students’ selected numbers between 1 and 10 were tabulated, and the relative frequencies calculated. The expected frequency for each number, assuming a uniform distribution, is total responses divided by 10.

Calculations of the relative frequencies reveal deviations from uniformity. Using the Chi-Square Goodness-of-Fit test, we compare observed frequencies to expected frequencies. The hypotheses are:

H0: The distribution of numbers is uniform.

Ha: The distribution is not uniform.

Calculations involve:

χ² = Σ [(O - E)² / E]

Where O = observed frequency, E = expected frequency.

Suppose, for example, the observed frequencies for some numbers are 15, 20, 10, etc., and the total responses sum to 100, implying E = 10 for each number.

Calculating the χ² statistic and comparing it with a critical value (from χ² distribution table with 9 degrees of freedom at α=0.05, which is approximately 16.92), we determine whether to reject H0.

If the computed χ² exceeds 16.92, we reject H0, concluding the distribution differs significantly from uniform. Otherwise, we cannot reject the null hypothesis.

This analysis reveals insights into whether students’ choices were truly random or influenced by biases or patterns, aligning with the article’s discussion on the complexity of generating random numbers.

Conclusion

This statistical investigation demonstrates how descriptive and inferential methods can be employed to analyze survey data. The shape analysis of distribution provides insight into data skewness, while confidence intervals quantify uncertainty around sample estimates. Hypothesis tests, including one-proportion z-tests, t-tests, and chi-square goodness-of-fit, enable evaluation of claims and underlying assumptions. Proper application of these techniques allows researchers to draw meaningful conclusions about student behaviors and preferences, informing educational strategies and further research opportunities.

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