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| Section | Weight | Objectives |
|---|---|---|
| Basic Numeracy & Algebra | 15% | - Exponents, roots, and basic formulas - Linear equations, inequalities, graphing functions - Arithmetic operations, fractions, decimals, percentages |
| Descriptive Statistics | 25% | - Measures of center: mean, median, mode - Measures of spread: range, IQR, variance, standard deviation - Graphical displays: histograms, boxplots, scatterplots - Types of data: categorical, discrete, continuous |
| Inferential Statistics & Study Design | 20% | - Sampling methods and bias - Confidence intervals for means/proportions - Hypothesis testing framework and interpretation - Observational studies vs experiments |
| Correlation & Regression | 20% | - Interpreting slope, intercept, and R-squared - Predictions and limitations of regression - Correlation coefficient and interpretation - Simple linear regression models |
| Probability Concepts | 20% | - Discrete and continuous probability distributions - Conditional probability and Venn diagrams - Normal distribution and empirical rule - Probability rules, independent and dependent events |
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NEW QUESTION # 120
A t-distribution is used when:
Answer: A
Explanation:
A t-distribution is used for inference about a population mean when the population standard deviation # is unknown, especially with a small sample. In that situation, the sample standard deviation s must estimate #, creating extra uncertainty. The t-distribution accounts for this added uncertainty with heavier tails than the standard normal distribution. As sample size increases, the t-distribution becomes closer to the normal distribution. Option B describes a z-procedure setting, where # is known and the normal distribution may be used. Option C is incorrect because categorical data are usually analyzed with proportions or chi-square procedures, not t-distributions for means. Option D is incomplete: t-procedures typically require quantitative continuous or approximately continuous data, but the decisive condition in the answer set is unknown # with a small sample. Study Guide references/topics: t-distribution, unknown standard deviation, small samples, inference for means.
NEW QUESTION # 121
Variance formula = ?
Answer: B
Explanation:
The sample variance formula is s² = #(x# # x#)² / (n # 1). In this formula, x# represents each data value, x# is the sample mean, and n is the sample size. The expression x# # x# gives each value's deviation from the mean. Squaring the deviations prevents negative and positive deviations from canceling and emphasizes larger departures. Dividing by n # 1 gives the sample variance, using degrees of freedom to correct bias when estimating population variance from a sample. Option B is the formula for the sample mean, not variance.
Option C squares the mean and does not measure spread. Option D resembles part of the standard deviation process but omits division by n # 1 and is not the variance formula. Variance is foundational because standard deviation is the square root of variance. Study Guide references/topics: variance, sample variance formula, mean deviations, measures of spread.
NEW QUESTION # 122
A commuter has a .4 probability of taking the bus to work, a .4 probability of driving a car, and a .2 probability of cycling. The probability of being late is .15 when taking the bus, .05 when driving a car, and .1 when cycling.
What is the probability of taking the bus and being on time, or driving a car and being on time?
Answer: C
Explanation:
This item requires multiplying conditional probabilities and then adding mutually exclusive outcomes.
"Taking the bus and being on time" means the commuter takes the bus and is not late. Since the probability of being late by bus is .15, the probability of being on time by bus is 1 # .15 = .85. Therefore, P(bus and on time)
= .4 × .85 = .34. For driving, the probability of being late is .05, so the probability of being on time is 1 # .05
= .95. Therefore, P(car and on time) = .4 × .95 = .38. Because a commuter cannot both take the bus and drive a car on the same trip, the events are mutually exclusive. Add the two joint probabilities: .34 + .38 = .72. The cycling information is not used because the question asks only about bus-on-time or car-on-time outcomes.
References/topics from the Study Guide: probability rules, complements, conditional probability, mutually exclusive events.
NEW QUESTION # 123
Determine z in the equation:
1/(6z) = 2/9
Answer: D
Explanation:
The equation 1/(6z) = 2/9 is a proportion because it sets two fractions equal to each other. To solve it, use cross multiplication. Multiply the numerator of the first fraction by the denominator of the second fraction, and multiply the denominator of the first fraction by the numerator of the second fraction: 1 × 9 = 2 × 6z. This gives 9 = 12z. To isolate z, divide both sides by 12: z = 9/12. This fraction can also be simplified to 3/4, but the answer choices list the unsimplified equivalent form 9/12. The other options do not satisfy the original equation. For example, if z = 2/54, then 6z is very small and the left side becomes much larger than 2/9. If z =
12/9 or 18/2, the denominator becomes too large, making the left side too small. References/topics from the Study Guide: proportions, solving equations, cross multiplication, equivalent fractions.
NEW QUESTION # 124
Standard error decreases when:
Answer: D
Explanation:
The standard error of the sample mean is calculated as SE = s/#n, where s is the sample standard deviation and n is the sample size. Since n appears in the denominator, increasing the sample size reduces the standard error, assuming the standard deviation stays constant. This reflects the fact that larger samples produce more stable estimates of the population mean. Option B is incorrect because decreasing sample size increases standard error. Option C is incorrect because a larger standard deviation increases standard error by increasing the numerator. Option D is incorrect because the mean affects the center of the distribution but does not directly determine the standard error. Standard error is about precision: smaller standard error means less sampling variability and a more precise estimate. That is why larger samples are preferred in inference. Study Guide references/topics: standard error, sample size, standard deviation, sampling variability.
NEW QUESTION # 125
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