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This question paper must be returned. Candidates are not permitted to remove any part of it from the examination room. Other instructions include writing your Student ID, Name, Surname, and Table Number at the top, placing your Macquarie University Campus Card prominently, following all invigilator instructions, and only using authorized materials during the exam. Non-programmable calculators are permitted, and students may bring one handwritten A4 sheet into the exam. No dictionaries are allowed.

The exam consists of five questions, totaling 75 marks, including short answer and calculation questions, covering content from the unit WSTA150 Business Statistics. All answers must be written in blue or black pen, with additional space available on the last pages. The exam duration is 2 hours plus 10 minutes reading time.

Paper For Above instruction

The following paper presents a comprehensive analysis of several statistical methods applied in various contexts, including probability distributions, hypothesis testing, regression analysis, and chi-squared tests. The questions require interpreting Excel outputs, performing statistical calculations, and critically analyzing the results within a real-world framework.

Introduction

Statistics plays a vital role in supporting decision-making across diverse fields such as agriculture, healthcare, sports, and public policy. The application of statistical tools allows researchers and practitioners to interpret data accurately, test hypotheses, and make predictions. In this paper, we explore several typical statistical analyses through problems derived from different scenarios, illustrating how statistical concepts are employed to draw meaningful conclusions.

Question 1: Probabilities and Hypothesis Testing in Plant Germination

The first question involves analyzing the germination time of plant seeds. The germination duration for a particular plant variety is modeled as a normally distributed random variable with a mean of 15 days and a standard deviation of 4 days. Using Excel functions, such as NORM.DIST and NORM.INV, we explore various probability questions and hypothesis testing related to the germination process.

i) The probability that seeds germinate within 19 days is calculated as P(X ≤ 19). Applying the Excel function NORM.DIST(19, 15, 4, TRUE), yields approximately 0.8413, which indicates an 84.13% chance that a seed will germinate in 19 days or less. Graphically, shading the area under the bell curve to the left of 19 days visually depicts this probability.

ii) The probability that a seed takes 12 or more days to germinate is P(X ≥ 12), calculated as 1 - P(X

iii) To find the probability that germination time lies between 12 and 18 days, we compute P(12 ≤ X ≤ 18) = P(X ≤ 18) - P(X ≤ 12). Using the corresponding Excel functions, P(X ≤ 18) ≈ 0.7733 and P(X ≤ 12) ≈ 0.2266, thus, the probability is roughly 0.5467. Shading the area between these points on the bell curve shows this likelihood.

iv) The day by which three-fourths of the seeds have germinated corresponds to the 75th percentile. Using NORM.INV(0.75, 15, 4), the result is approximately 17.70 days, indicating that about 75% of the seeds are expected to germinate by this time.

v) Testing whether 60% of seeds germinate by day 15 involves a hypothesis test for proportions. Based on a sample of 50 seeds with 33 germinated, this gives a sample proportion of 0.66. The null hypothesis H0: p = 0.60 is tested against the alternative H1: p ≠ 0.60. Computation of the Z-statistic and corresponding p-value indicates whether there is significant evidence to reject H0. Using the normal approximation, Z = (0.66 - 0.60) / sqrt(0.6 * 0.4 / 50) ≈ 1.00. Since this Z-value corresponds to a p-value greater than 0.05, there is not enough evidence to reject the null hypothesis, suggesting the proportion is consistent with 60%.

Question 2: Testing Average Waiting Times for Cancer Surgery

In this scenario, the administration claims an average waiting time of 16 days for cancer surgeries across Australian hospitals. An independent sample involved 36 patients with a sample mean of 15.42 days and a standard deviation of 6 days. The application of a t-test evaluates whether the true mean significantly differs from 16 days.

a) The most appropriate test is a one-sample t-test, because the population standard deviation is unknown and the sample size is moderate (n=36). This method compares the sample mean against the claimed population mean, considering sample variability.

b) The t-test statistic is calculated as t = (sample mean - hypothesized mean) / (sample SD / sqrt(n)) = (15.42 - 16) / (6 / sqrt(36)) ≈ -0.58. Using Excel’s T.DIST.2T function or statistical software provides the p-value. For t = -0.58 with df=35, the p-value exceeds 0.05, indicating no significant difference at 5% significance level. Thus, there is insufficient evidence to refute the claim that the mean waiting time is 16 days.

Constructing a 95% confidence interval involves calculating the margin of error: ME = t(0.975,35) (6/√36) ≈ 2.030 1, which is approximately 2.03 days. The confidence interval is from 15.42 - 2.03 ≈ 13.39 days to 15.42 + 2.03 ≈ 17.45 days. The interval encompasses 16 days, aligning with the claim that the average waiting time is around 16 days.

Question 3: Comparing Cancer Surgery Waiting Times in NSW and Victoria

The research question investigates whether there is a difference between average waiting times for cancer surgeries in New South Wales and Victoria. The two tests provided include a two-sample assuming equal variances and a paired sample test. Since the data samples are independent, the most appropriate is the two-sample t-test assuming equal variances.

a) For independent datasets with different groups, Test 1 (the two-sample t-test assuming equal variances) is more appropriate because the samples are independent and the goal is to compare their means.

b) The t-statistic is approximately 1.2323 with a p-value of about 0.222, which exceeds the significance threshold of 0.05. Therefore, there is no statistically significant difference between the mean waiting times in NSW and Victoria based on this data, indicating that the two populations could have similar average waiting times.

Question 4: Analyzing Class Size and Academic Performance

This question involves regression analysis, examining the relationship between class size and students' average marks. The regression output indicates a strong negative correlation (R = 0.9574) and a high coefficient of determination (R² = 0.9167), signifying that class size explains a significant proportion of variability in academic performance.

a) The scatter plot indicates a strong inverse relationship: as class size increases, average marks tend to decrease. The data points are tightly clustered around the regression line, confirming the strength of this relationship.

b) The residual plot appears randomly scattered around zero without discernible patterns, suggesting the residuals are approximately normally distributed, fulfilling the normality assumption.

c) The assumption of constant variance (homoscedasticity) can be examined through the residual plot. Uniform spread of residuals across the predictor variable indicates this assumption holds. If residuals fan out or contract, heteroscedasticity may be present.

d) The correlation coefficient, already given, is positive or negative with a magnitude of approximately 0.9574, indicating a very strong negative correlation, meaning larger class sizes are associated with lower average marks.

e) The coefficient of determination (R² = 0.9167) indicates about 91.67% of the variation in student performance can be explained by class size, highlighting the importance of class size in influencing academic achievement.

f) The least squares regression equation is: Predicted Average Marks = 91.1918 - 1.0568 * (Class Size). It suggests that each additional student in a class reduces the average mark by approximately 1.0568 points.

g) For a class size of 20, substituting into the regression equation yields: 91.1918 - 1.0568 * 20 ≈ 91.1918 - 21.136 ≈ 70.06. Therefore, the predicted average mark is approximately 70.06 points.

h) For a class size of 3, the predicted average is 91.1918 - 1.0568 * 3 ≈ 91.1918 - 3.1704 ≈ 88.02.

i) To test whether class size significantly affects average marks, an F-test or t-test on the regression coefficient is performed. The significance of the slope (p-value

Question 5: Chi-Squared Test for Medal Distribution

The final question pertains to the chi-squared goodness-of-fit test, evaluating whether the observed medal counts align with the predicted proportions for each country at the 2018 Winter Olympics. The null hypothesis states that the distribution matches the expected proportions, while the alternative suggests deviations exist. The p-value provided (0.029) indicates there is statistically significant evidence to reject the null hypothesis at the 5% level, meaning the actual medals won differ significantly from the predicted distributions. This highlights potential inconsistencies or unpredictable factors influencing medal outcomes beyond initial forecasts.

Conclusion

Overall, these analyses demonstrate the importance of statistical methods in decision-making across disciplines—from evaluating plant germination probabilities and healthcare wait times to assessing educational outcomes and sporting event predictions. Employing hypothesis testing, regression, and goodness-of-fit tests allows researchers to draw credible inferences, support policy decisions, and understand underlying relationships.

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