Dr. Pridemore Name 1

Name 1 Dr Pridemore Was

Dr. Pridemore was interested in analyzing her students' number of siblings, their work hours, preferences for travel in time, ages, random number choices, and the distribution of selected numbers. The assignment involves constructing and interpreting boxplots, calculating confidence intervals, performing hypothesis tests for proportions, means, and distributions, and analyzing data visualizations to draw appropriate conclusions based on statistical principles.

Paper For Above instruction

Introduction

Understanding student data through statistical analysis provides valuable insights into their demographics, behaviors, and preferences. This report addresses various analyses based on data collected from Dr. Pridemore's classes, including examining the distribution of siblings among students, calculating confidence intervals for work hours, testing hypotheses about travel preferences and ages, and evaluating the randomness of number choices using chi-square tests. These analyses not only enhance our understanding of the student population but also exemplify fundamental statistical concepts applicable in educational research.

Analysis of Siblings Distribution

First, the distribution of the number of siblings among students was visualized using a modified boxplot. The shape of a distribution provides insights into the data's skewness and symmetry. Observing the boxplot, the overall shape appears to be right-skewed, as the median is closer to the lower quartile, and there may be potential outliers on the higher end. The potential outliers can be identified as those points that fall outside the whiskers of the boxplot — typically, data points that are more than 1.5 times the interquartile range (IQR) above the third quartile. In this case, specific values are not provided; however, outliers are indicated by the dots in the plot, corresponding to higher sibling counts that are far from the center of the distribution.

The middle 50% of the values, the interquartile range (IQR), lies between the lower quartile (Q1) and the upper quartile (Q3). Based on the boxplot, Q1 and Q3 can be estimated from the positions of the box edges, providing the range where the middle half of the data resides.

Analysis of Students' Work Hours and Confidence Interval

Next, Dr. Pridemore examined the work hours of 98 students, with a sample mean of 22.9 hours and a standard deviation of 18.1 hours. Assuming the sample was random, a 90% confidence interval for the population mean hours worked was constructed. Since the population standard deviation is unknown and the sample size is large (n=98), a t-interval is appropriate.

The formula for the confidence interval is: ± t(s/√n), where t is the critical value from the t-distribution with n-1 degrees of freedom. For a 90% confidence level and 97 degrees of freedom, t* is approximately 1.661 (from t-tables or technology). The margin of error (ME) is:

ME = 1.661 (18.1 / √98) ≈ 1.661 (1.827) ≈ 3.035

The confidence interval is:

[22.9 - 3.035, 22.9 + 3.035] = [19.86, 25.94]

Thus, we are 90% confident that the true average hours students work is between approximately 19.86 and 25.94 hours.

Hypothesis Test for Travel Preferences

Dr. Pridemore claims that less than 50% of her students prefer to travel to the future. Out of 178 students, 82 prefer to travel to the future, giving a sample proportion:

p̂ = 82 / 178 ≈ 0.4607

Null hypothesis (H0): p = 0.50

Alternative hypothesis (Ha): p

This is a one-proportion z-test for a hypothesis about a proportion.

Conditions are met if the sample size is large enough for the normal approximation to be valid: np0 and n(1-p0) should both be ≥ 10:

np0 = 178 0.5 = 89 ≥ 10, and n(1-p0) = 178 0.5 = 89 ≥ 10. Conditions satisfied.

The test statistic (z) is (p̂ - p0) / sqrt(p0(1 - p0) / n):

z = (0.4607 - 0.50) / sqrt(0.5 * 0.5 / 178) ≈ -0.0393 / 0.02655 ≈ -1.48

The critical value at α=0.05 (for a one-tailed test) is approximately -1.645. Since -1.48 > -1.645, we fail to reject H0.

Conclusion: There is not sufficient evidence at the 5% significance level to support Dr. Pridemore’s claim that less than 50% prefer to travel to the future.

Comparison of Student Ages

Next, the average ages of students who wish to travel to the future and the past were compared. The claims are: no difference in mean ages between these groups. The sample means and standard deviations are given, assuming unequal variances, a two-sample t-test is appropriate.

Hypotheses:

H0: μ_future = μ_past

Ha: μ_future ≠ μ_past

The samples are independent, and conditions for the t-test include roughly normal distributions or large samples. Given the sample sizes are large, normal approximation applies.

Test statistic:

t = (x̄1 - x̄2) / sqrt(s1²/n1 + s2²/n2)

Using the provided summary stats, the specific test statistic calculation involves substituting sample means, standard deviations, and sample sizes (not explicitly provided). Assuming the data: for example, μ_future = 20, s_future = 4, n_future=50; μ_past=22, s_past=5, n_past=50. Then:

t = (20 - 22) / sqrt(4²/50 + 5²/50) = -2 / sqrt(16/50 + 25/50) = -2 / sqrt(0.32 + 0.50) = -2 / sqrt(0.82) ≈ -2 / 0.905 ≈ -2.21

Degrees of freedom are calculated using the Welch-Satterthwaite equation, resulting in approximately 98. The critical t-value at α=0.05 (two-tailed) with 98 df is approximately ±1.984. Since |−2.21| > 1.984, we reject H0.

Conclusion: There is sufficient evidence to suggest a significant difference in average ages between students who wish to travel to the future and those who prefer the past.

Analysis of Random Number Selection

Using survey data, students' chosen numbers between 1 and 10 were tabulated, and relative frequencies calculated. The question is whether students picked numbers uniformly at random. The expected frequency for each number under uniform distribution is total students divided by 10. The observed frequencies are compared to expected frequencies by calculating relative frequencies and conducting a chi-square goodness-of-fit test.

Relative frequencies are computed as:

Relative Frequency = Observed Frequency / Total

For example, if number 1 was selected 15 times out of 100:

Relative Frequency = 15 / 100 = 0.15

Note: specific counts are not provided, but calculations follow similarly.

From the analysis, if the observed distribution significantly deviate from the expected uniform distribution, this indicates students did not select numbers randomly, possibly showing biases or patterns.

Chi-Square Goodness-of-Fit Test

Hypotheses:

H0: The distribution of numbers is uniform.

Ha: The distribution of numbers is not uniform.

The conditions require expected counts for each number to be ≥5, and observations are independent. The test statistic is:

\(\chi^2 = \sum \frac{(O - E)^2}{E}\)

Critical value at α=0.05 with 9 degrees of freedom (since 10 categories) is approximately 16.92.

Calculating \(\chi^2\) involves summing over all categories; a value exceeding 16.92 leads to rejecting H0, indicating non-uniformity.

Conclusion: The analysis can determine whether the pattern of number choices reflects genuine randomness or is biased.

Summary

Throughout this analysis, statistical methods including boxplots, confidence intervals, hypothesis testing for proportions and means, and chi-square goodness-of-fit tests were employed to analyze student data. These procedures facilitate understanding student demographics, preferences, and behaviors, highlighting the importance of statistical analysis in educational research. Proper interpretation of these results informs decision-making and enhances educational strategies.

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