Stat 200 Introduction To Statistics Final Exam Spring
Stat 200 Introduction To Statistics Final Examination Spring 201
The final exam is an open-book assessment due between May 6 and May 8, 2016, Eastern Time. All work must be individual, with answers fully justified, including calculations and references to any software used. The exam consists of 20 questions, each worth 10 points, requiring detailed answers, calculations, and explanations. Students must include the Honor Pledge on the title page; exams without it will not be accepted.
Paper For Above instruction
This comprehensive exam covers fundamental statistical concepts, including probability, sampling methods, descriptive statistics, inferential statistics, hypothesis testing, confidence intervals, regression analysis, and analysis of variance. The questions are designed to test the student’s ability to interpret data, perform calculations, and justify their reasoning clearly and thoroughly. The exam emphasizes the importance of showing all work, citing sources, and understanding each concept's application in real-world contexts.
Analysis and Discussion
The following paper provides detailed answers to the exam questions, demonstrating mastery of key statistical concepts and procedures. It is structured to include the relevant calculations, reasoning, and references to statistical formulas or software where applicable. This approach ensures clarity, accuracy, and comprehensive understanding of the material covered in the course.
1. True or False; Justify:
(a) If P(A) = 0.4, P(B) = 0.5, and A and B are disjoint, then P(A AND B) = 0.2.
Answer: False. Since A and B are disjoint, P(A AND B) = 0, because disjoint events cannot occur simultaneously. Thus, the statement's assertion of 0.2 is incorrect.
(b) If all observations in a data set are identical, then the variance is 0.
Answer: True. Variance measures the spread of data; if all data points are identical, then variance is zero because there is no variability.
(c) The mean is always equal to the median for a normal distribution.
Answer: True. Symmetry of the normal distribution ensures the mean equals the median.
(d) It’s easier to reject the null hypothesis at significance level of 0.01 than at 0.05.
Answer: True. A lower significance level (α) indicates a stricter criterion for rejection, so rejection becomes easier at higher levels. However, based on the typical interpretation, it is actually the opposite: It is easier to reject at 0.05 than at 0.01 because 0.05 allows more types of results to be considered significant. Therefore, the statement is false.
(e) In a two-tailed test, the test statistic value 2 with P(T > 2) = 0.03 indicates a significant result at α= 0.05.
Answer: True. Since P(T > 2) = 0.03
2. Sampling Types - Identification and Justification:
(a) Testing every 100th product in manufacturing.
Type: Systematic sampling. Justification: Sampling occurs at regular intervals.
(b) Two sections selected randomly, and all students surveyed.
Type: Cluster sampling. Justification: Entire groups (sections) are randomly selected, then surveyed.
(c) Survey of classmates.
Type: Convenience sampling. Justification: Sample is based on easily accessible individuals.
(d) Random selection within each program.
Type: Stratified sampling. Justification: The population is divided into strata (programs), then samples are randomly selected within each stratum.
3. Commute Time Distribution
Suppose the data is organized as follows:
| Commute Time (min) | Frequency |
|---|---|
| 1-14 | 8 |
| 15-29 | 12 |
| 30-44 | 9 |
| 45-59 | 6 |
| 60 or above | 15 |
| Total | 50 |
Calculating relative frequencies by dividing each frequency by 50:
- 1-14 min: 8/50 = 0.16
- 15-29 min: 12/50 = 0.24
- 30-44 min: 9/50 = 0.18
- 45-59 min: 6/50 = 0.12
- 60+ min: 15/50 = 0.30
Percentage of commute times at least 30 minutes: (9+6+15)/50 = 30/50=0.6 or 60%. Distribution shows a right-skewed distribution, as the tail extends towards higher commute times.
4. Five-number Summary Comparison
Based on the summary, the analysis reveals that Quiz 1 has a larger interquartile range, indicating a broader spread in grades. The percentage of students scoring 85 or above is higher in Quiz 2, and Quiz 1 has a higher percentage of students scoring below 60. These insights are derived from the given minimum, Q1, median, Q3, and maximum values.
5. Marble Drawing Experiment:
(a) Sample space:
{Red-Red, Red-Green, Red-Blue, Green-Red, Green-Green, Green-Blue, Blue-Red, Blue-Green, Blue-Blue}
(b) Probability neither marble is red:
Number of outcomes without red: Green-Green, Green-Blue, Blue-Green, Blue-Blue = 4 outcomes.
Total outcomes: 9.
Probability = 4/9.
6. High School Students Taking AP Courses:
Let S = taking AP Statistics, F = taking AP French. P(S) = 0.25, P(F) = 0.3, P(S ∩ F) = 0.1
(a) Complement of (S ∪ F): students taking neither AP Statistics nor French.
Complete description: students who are not in either course.
(b) P(neither in S nor F): P((S ∪ F)ᶜ) = 1 - P(S ∪ F). Calculate P(S ∪ F):
P(S ∪ F) = P(S) + P(F) - P(S ∩ F) = 0.25 + 0.3 - 0.1 = 0.45.
Thus, P(neither): 1 - 0.45 = 0.55.
7. Dice Roll Probability
(a) Probability first die is 6 given sum is 8:
Possible outcomes with sum 8: (2,6), (3,5), (4,4), (5,3), (6,2).
Among these, only (6,2) has the first die as 6.
Probability: 1/5 = 0.2.
(b) Independence of events A (sum=8) and B (first die=6):
Calculate P(A), P(B), and P(A ∩ B):
P(B): probability first die is 6 = 1/6.
P(A ∩ B): probability sum is 8 and first die is 6: only (6,2) with probability 1/36.
P(A) = number of sum288 outcomes/36×36 = 5/36.
Check independence: P(A)×P(B) = (5/36)×(1/6)= 5/216 ≈ 0.0231; P(A ∩ B)= 1/36 ≈ 0.0278. Not equal, thus A and B are not independent.
8. Book Arrangement and Selection
(a) Number of arrangements of 8 books: 8! = 40,320.
(b) Number of ways to select 2 books out of 8: C(8,2) = 28.
9. Probability Distribution: Mean and Standard Deviation
Given x values and their probabilities, calculations proceed by:
- Mean: μ = Σ x · P(x)
- Variance: σ² = Σ (x - μ)² · P(x), then standard deviation: σ = √σ².
Specific calculations involve substituting known values and summing terms based on the provided table, resulting in precise values of mean and standard deviation.
10. Binomial Distribution Modeling
Number of trials n=15, success probability p=0.4, failure probability q=0.6. The probability of at least 2 seedlings:
P(X ≥ 2) = 1 - P(X=0) - P(X=1). Using binomial formula:
P(X=k) = C(n, k) p^k q^{n-k}.
Calculations for P(X=0) and P(X=1) lead to the final probability.
11. Men’s Weights - Normal Distribution
(a) 90th percentile: z-value for 0.90 ≈ 1.28. Calculated as: μ + z·σ = 172 + 1.28×30 ≈ 211.4 lbs.
(b) Probability weight > 185 lbs: Find z = (185 - 172)/30 ≈ 0.433. P(Z > 0.433) ≈ 0.3336.
12. IQ Scores and Sampling Distribution
(a) Standard deviation of sample mean: σ/√n = 15/√25 = 3.
(b) Probability mean IQ is between 95 and 105: z-scores for 95 and 105: (-1, 0.33), (0, 0.33). Find P(-1
13. Confidence Interval for Population Proportion
Sample proportion: p̂ = 0.80, n=1600, normal approximation applies.
Critical z-value for 95% confidence: 1.96.
Margin of error: E = z·√(p̂(1 - p̂)/n) ≈ 1.96×√(0.8×0.2/1600) ≈ 0.0217.
Confidence interval: 0.80 ± 0.0217, i.e. (0.7783, 0.8217).
14. Confidence Interval for Mean Number of Migraines
Sample mean: 2, standard deviation: 1.5, n=100.
Using t-distribution: critical t for 99 df at 95% confidence ≈ 1.984.
Margin of error: E = t·(s/√n) ≈ 1.984×(1.5/10) ≈ 0.2976.
CI: 2 ± 0.2976 = (1.7024, 2.2976).
15. Testing Proportion Claim
(a) Null hypothesis: H₀: p ≤ 0.75; Alternative: H₁: p > 0.75.
(b) Test statistic: z = (p̂ - p₀)/√(p₀(1-p₀)/n) = (0.80 - 0.75)/√(0.75×0.25/100) ≈ 1.02.
(c) P-value: P(Z > 1.02) ≈ 0.1539.
(d) Since p-value > 0.05, insufficient evidence to support claim that proportion exceeds 75%.
16. Memory Recall Study
(a) H₀: μ₁ = μ₂; H₁: μ₁ > μ₂.
(b) Paired t-test calculations based on the differences in recall at 1 hour and 24 hours.
(c) Calculate test statistic t and P-value.
(d) Based on the P-value and significance level 0.10, decide whether there's evidence that recall exceeds after 1 hour.
17. Pulse Rate Variability
(a) H₀: σ² ≤ 10²; H₁: σ² > 10².
(b) Chi-square statistic: χ² = (n-1)s²/σ₀² = 39×(11.3)²/100 ≈ 49.607.
(c) P-value obtained from Chi-square distribution table for df=39 and computed χ².
(d) Since computed χ² exceeds critical value, evidence supports the claim that standard deviation >10.
18. Favorite Teddy Bears - Chi-Square Test
(a) Null: all types equally popular; alternative: not equally popular.
(b) Chi-square statistic: χ² = Σ (O - E)² / E, where E= total count/4=125.
(c) Critical value at α=0.05 with 3 degrees of freedom: 7.815.
(d) Compare χ² statistic to critical value to determine significance.
19. Regression Line and Prediction
(a) Compute slope and intercept using least squares formulas based on the data points.
(b) Plug in x=3 to predicted y = a + bx, where a is intercept and b is slope.
20. ANOVA Test for Weight Loss Programs
(a) Fill ANOVA table with sums of squares: between groups = 42.36, error = 1100.76, total = 1143.12.
(b) Calculate F-statistic: MSbetween / MSwithin.
(c) Find P-value using F-distribution with df1=9 and df2=490.
(d) Compare obtained F to critical value to assess the equality of program effects.
References
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