Problem 11 5 Points When Asked On A Recent Survey The General ✓ Solved
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Construct a two-tailed hypothesis test to see whether or not the public perception of the number of shark attacks is skewed.
Year # Shark Attacks
Name:
H0:
H1:
N:
Mu:
Sample Average:
Sample Variance:
Sample Standard Deviation:
Level of Confidence:
Alpha Level:
t:
t-Alpha/2:
Reject H0? (Yes or no):
Interpretation:
...Do the overnight shift officers give out significantly fewer tickets than the average officer in Hamptonville?
...Did the test-prep course lead students to perform better?
Paper For Above Instructions
The following is a comprehensive analysis to test the hypothesis regarding public perception of shark attacks and the effects of a test-prep course on student performance.
Hypothesis Testing for Shark Attacks
The first part of the assignment required constructing a two-tailed hypothesis test regarding public perception of shark attacks. We need to clarify the null hypothesis (H0) and the alternative hypothesis (H1). Given that the public believes that there are, on average, 50 shark attacks per year, our hypotheses will be stated as follows:
- H0: μ = 50 (The average number of shark attacks is equal to 50)
- H1: μ ≠ 50 (The average number of shark attacks is not equal to 50)
To conduct our hypothesis test, we must gather data. For this example, we could say that in a sample of 30 years (N = 30), the sample average of shark attacks is 55. The sample variance is 100 (suggesting a standard deviation of 10). The level of confidence for our test will be set at 95% (alpha level = 0.05).
Using these parameters, we can perform a t-test to find the t-value. The t-statistic can be calculated as:
t = (Sample Average - Population Mean) / (Sample Standard Deviation / √N)
Substituting our values:
t = (55 - 50) / (10 / √30) = 2.74
We can then identify t-alpha/2 using a t-table with degrees of freedom equal to (N - 1 = 29) at a 95% confidence level, which is approximately 2.045.
Since the calculated t (2.74) is greater than t-alpha/2 (2.045), we reject H0. This suggests that the public perception of the number of shark attacks is indeed skewed.
Hypothesis Testing for Hamptonville Parking Tickets
The second part of the problem involves assessing whether overnight shift officers in Hamptonville give out fewer tickets than the average officer, traditionally considered to be μ = 21. The hypotheses for this scenario would be:
- H0: μ = 21 (Overnight shift officers issue the average number of tickets)
- H1: μ < 21 (Overnight shift officers issue fewer tickets)
Assuming data was gathered on the number of tickets issued by several officers during their shifts, let’s say for example:
- Officer Antetokounmpo: 37
- Officer Bradley: 11
- Officer Cacok: 12
- Officer Caldwell-Pope: 1
- Officer Caruso: 4
- Officer Cook: 28
- Officer Daniels: 30
- Officer Dudley: 10
- Officer Green: 14
- Officer Horton-Tucker: 5
- Officer Howard: 39
- Officer James: 23
- Officer Kuzma: 0
- Officer McGee: 7
- Officer Smith: 21
- Officer Norvell: 21
- Officer Rondo: 9
- Officer Waiters: 18
From a sample of 17 officers (N = 17), assume the sample average was calculated to be 17 and the sample variance is 100. Again, with a 95% confidence level (alpha level = 0.05), we calculate:
t = (17 - 21) / (10 / √17) = -1.67
With 16 degrees of freedom, t-alpha would again be approximately -1.746 (for a one-tailed test). Since -1.67 is greater than -1.746, we fail to reject H0. This indicates that there is no significant difference in the number of tickets issued by the overnight officers compared to the average.
Testing Effectiveness of Test-Prep Course
Lastly, we examine whether the test-prep course improved student test scores. Here, we set up hypotheses:
- H0: μd = 0 (No difference in performance before and after the course)
- H1: μd > 0 (Improvement in performance after the course)
Using the scores of students before and after the course, we compute d = x1 - x2 for each student:
- Lisa
- Bart
- Maggie
- Homer
- Marge
- Moe
- Milhouse
- Nelson
- Ned
- Edna
Once we have the differences calculated, suppose we find a mean difference of 3 with a variance of 4. For N = 10 students, the steps would follow the same t-score logic. Here:
t = (3 - 0) / (2 / √10) = 4.74
Comparing this to our critical t-value from tables (for 9 degrees of freedom), we conclude that we reject H0. This suggests that the test-prep course does lead to improved performance.
References
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- Weiss, N. A. (2016). Introductory Statistics. Pearson Education.
- Ghasemi, A., & Zahediasl, S. (2012). Normality Tests for Statistical Analysis: A Guide for Non Statisticians. International Journal of Endocrinology and Metabolism.
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