University:
University of the People
Course:
MATH 1281 | Statistical Inference
Academic year:

2024
Views:

455

Pages:

5
Author:

amrinderjeet kaur

UNIT 3 LEARNING JOURNAL Part 1 Barking Deer Foraging Preferences a. Hypotheses Null Hypothesis (H0H_0H0): Barking deer do not have a predilection for any one habitat; instead, they browse in a variety of habitats according to their availability. Alternative Hypothesis (HaH_aHa): The observed frequencies differ greatly from the expected frequencies because barking deer favor certain areas over others for foraging. b. Type of Test We compare the observed foraging site frequencies with the expected frequencies based on habitat availability using a Chi-Square Goodness-of-Fit Test. c. Assumptions & Conditions For the Chi-Square Goodness-of-Fit test to be valid: 1. Independence: Every foraging location is separate from the others. If the data were sampled at random, this assumption would be true. 2. Expected Counts: There should be at least five expected counts for each. Below, we'll examine this. d. Hypothesis Test Step 1: Compute Expected Counts The expected counts are calculated as: E=(Total Foraging Sites)×(Proportion of Habitat) Check Expected Count Condition ● The criterion is satisfied since all categories' predicted counts are higher than 5. Step 2: Compute Chi-Square Statistic ≈ 372.41 Step 3: Compute Degrees of Freedom df = Number of categories − 1 = 4 − 1 = 3 Step 4: Conclusion The p-value is significantly smaller than our significance level (α=0.05) because the problem specifies that it is <0.001. As a result, we reject the null hypothesis and come to the conclusion that barking deer forage in particular areas. Part 2 Testing for Imbalance in Experiment Groups a. Calculate the Actual Number of Site Visitors in Each Category We are given percentages for downloads and no downloads per position and the total number of site visitors (N=501). To find the actual number of visitors, we multiply the percentage (converted to a decimal) by the total visitors. Now, calculating the number of visitors for each response category: ● Position 1 (Download): 501×0.16=80.16≈80 ● Position 1 (No Download): 501×0.209=104.71≈105 ● Position 2 (Download): 501×0.148=74.15≈74 ● Position 2 (No Download): 501×0.212=106.31≈106 ● Position 3 (Download): 501×0.119=59.62≈60 ● Position 3 (No Download): 501×0.152=76.15≈76 b. Testing for Balance Across Experiment Groups Step 1: Define Hypotheses ● Null Hypothesis (H0): The three groups were assigned equally (i.e., each group should have roughly the same number of participants). ● Alternative Hypothesis (Ha): The groups were not assigned equally, meaning there is an imbalance. Step 2: Check Conditions for Chi-Square Test 1. Independence: Each visitor is randomly assigned to a position, so their placement is independent. 2. Expected Counts: The expected count for each group should be at least 5, which is true here. Step 3: Compute the Chi-Square Test Statistic The expected count for each position is: Now, compute the Chi-Square statistic: ≈ 8.71 Step 4: Compute Degrees of Freedom → df = Number of groups − 1 = 3 − 1 = 2 Step 5: Conclusion We reject the null hypothesis since the p-value = 0.01215 is less than the significance level (α=0.05). This indicates that there is proof that the visitors were not evenly distributed among the groups. REFERENCES: Moore, D. S., McCabe, G. P., Craig, B. A., & Alwan, L. C. (2021). The practice of statistics for business and economics (5th ed.). W. H. Freeman.