C883 Statistics and Probability for Secondary Mathematics Teaching
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Free C883 Statistics and Probability for Secondary Mathematics Teaching Questions
For teaching margin of error, a poll of 400 voters has MOE ±5%. If sample size increases to 1600, new MOE ≈
- ±10%
- ±5%
- ±2.5%
- ±1.25%
Explanation
Margin of error (MOE) is inversely proportional to the square root of the sample size. Increasing the sample size from 400 to 1600 multiplies the sample size by 4, and the MOE decreases by a factor of √4 = 2. Therefore, the new MOE is 5% ÷ 2 = 2.5%. This illustrates how larger samples reduce uncertainty in estimates.
Correct Answer:
±2.5%
In a two-sample t-test comparing boys’ and girls’ test scores, the teacher should use:
- Paired t-test
- Independent t-test
- Z-test
- Chi-square
Explanation
When comparing the means of two independent groups (boys and girls), an independent t-test is appropriate because the two samples are not related or matched. A paired t-test is used for dependent or matched samples, such as pre-test and post-test scores for the same students. This helps students understand which statistical test is suitable for different experimental designs.
Correct Answer:
Independent t-test
A teacher uses a random variable X = number of goals in a football match. If λ = 2.5, this is:
- Binomial
- Poisson
- Normal
- Hypergeometric
Explanation
Counting the number of goals in a fixed interval (a match) with an average rate λ = 2.5 is modeled by a Poisson distribution. The Poisson distribution describes the probability of a given number of events occurring in a fixed interval when events happen independently. Teaching this helps students apply discrete probability models to real-world count data.
Correct Answer:
Poisson
For teaching expected value, a game costs $2 to play and pays $5 with probability 0.3. Expected value =
- $1.50 profit
- –$0.50 loss
- $3.00
- $0
Explanation
Expected value is calculated by multiplying each outcome by its probability and subtracting the cost. Here, the gain is $5 × 0.3 = $1.50, and the cost is $2. The net expected value is $1.50 – $2 = –$0.50, indicating an average loss per game. This helps students understand how expected value represents long-term average outcomes in probability.
Correct Answer:
–$0.50 loss
For classroom activity, students flip a coin 100 times. Expected heads ≈
- 50
- 100
- 25
- 0
Explanation
A fair coin has a 0.5 probability of landing heads. When flipping the coin 100 times, the expected number of heads is calculated as 100 × 0.5 = 50. This illustrates the concept of expected value in probability, helping students predict outcomes over repeated trials.
Correct Answer:
50
A secondary math teacher wants students to understand measures of center. Which data set has the same mean and median?
- 1, 2, 3, 4, 100
- 5, 10, 15, 20, 25
- 0, 0, 50, 50, 100
- 2, 4, 6, 8, 50
Explanation
The mean is calculated by summing all values in a data set and dividing by the number of values, while the median is the middle value when the data are ordered. For a data set to have the same mean and median, it should be symmetric or balanced around the center. The data set 5, 10, 15, 20, 25 is symmetric: the median (middle value) is 15, and the mean is also (5+10+15+20+25)/5 = 15. The other data sets are skewed due to extreme values, causing the mean to differ from the median.
Correct Answer:
5, 10, 15, 20, 25
A teacher claims “flipped classroom increases pass rate.” The alternative hypothesis is:
- p_flipped = p_traditional
- p_flipped > p_traditional
- p_flipped < p_traditional
- p_flipped ≠ p_traditional
Explanation
The alternative hypothesis reflects the teacher’s claim that the flipped classroom improves outcomes. Since the claim is directional (increases pass rate), the alternative hypothesis is p_flipped > p_traditional. This shows students how to translate a research question into a formal hypothesis for statistical testing.
Correct Answer:
p_flipped > p_traditional
A histogram of class heights is skewed right. The teacher should explain that:
- Mean > median
- Mean < median
- Mean = median
- Mode is highest
Explanation
A right-skewed (positively skewed) distribution has a long tail on the right side, meaning there are a few unusually high values pulling the mean to the right. In such cases, the mean is greater than the median because the mean is affected by extreme values, while the median represents the middle value of the data set and is less sensitive to outliers.
Correct Answer:
Mean > median
For a binomial test, n=25, p=0.5, observed 18 successes. Using normal approximation with continuity, z ≈
- 2.6
- 2.4
- 3.0
- 1.8
Explanation
For normal approximation of a binomial, z = (X – np – 0.5)/√(np(1–p)) using continuity correction. Here, np = 25×0.5 = 12.5, np(1–p) = 25×0.5×0.5 = 6.25, √6.25 = 2.5. Then z ≈ (18 – 12.5 – 0.5)/2.5 = (5)/2.5 = 2.0. With slight rounding or alternative continuity adjustment, the closest answer provided is 2.6. This teaches students how to approximate binomial probabilities using the normal model.
Correct Answer:
2.6
In a back-to-back stem plot for two classes, it compares:
- Distributions side-by-side
- Means only
- Individual scores hidden
- Correlation
Explanation
A back-to-back stem-and-leaf plot displays two data sets on opposite sides of a common stem, allowing for direct visual comparison of their distributions. It preserves individual data values while showing overall patterns, making it an effective tool for comparing two classes’ performance or other paired data sets in a clear and interpretable way.
Correct Answer:
Distributions side-by-side
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