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| Section | Weight | Objectives |
|---|---|---|
| Topic 1: Basic Numeracy & Algebra | 15% | - Exponents, roots, and basic formulas - Arithmetic operations, fractions, decimals, percentages - Linear equations, inequalities, graphing functions |
| Topic 2: Correlation & Regression | 20% | - Interpreting slope, intercept, and R-squared - Simple linear regression models - Predictions and limitations of regression - Correlation coefficient and interpretation |
| Topic 3: Inferential Statistics & Study Design | 20% | - Observational studies vs experiments - Sampling methods and bias - Hypothesis testing framework and interpretation - Confidence intervals for means/proportions |
| Topic 4: Descriptive Statistics | 25% | - Types of data: categorical, discrete, continuous - Measures of spread: range, IQR, variance, standard deviation - Measures of center: mean, median, mode - Graphical displays: histograms, boxplots, scatterplots |
| Topic 5: Probability Concepts | 20% | - Normal distribution and empirical rule - Discrete and continuous probability distributions - Probability rules, independent and dependent events - Conditional probability and Venn diagrams |
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NEW QUESTION # 94
Sample mean = 60; point estimate of population mean = ?
Answer: A
Explanation:
A point estimate is a single sample statistic used to estimate an unknown population parameter. When estimating a population mean, the standard point estimate is the sample mean. In this question, the sample mean is given as 60. Therefore, the point estimate of the population mean is also 60. This does not mean the true population mean is guaranteed to equal 60; rather, 60 is the best single numerical estimate based on the available sample. Confidence intervals build on this point estimate by adding and subtracting a margin of error to reflect sampling uncertainty. Option B is incorrect because although the true population mean is unknown, the point estimate is known. Option C and option D do not correspond to the sample statistic provided. The logic is central to inferential statistics: use sample evidence to estimate population characteristics. Study Guide references/topics: sample mean, population mean, point estimation, inferential statistics.
NEW QUESTION # 95
One-sample t-test compares:
Answer: B
Explanation:
A one-sample t-test is used to compare a sample mean to a hypothesized or known population mean when the population standard deviation is unknown and the data are approximately normal or the sample size is sufficiently large. The test statistic evaluates how far the sample mean is from the hypothesized mean in standard error units. Option A is therefore correct. A two-sample t-test compares means from two independent groups, so option B describes a different test. Tests of variances use procedures such as chi-square or F-based methods depending on context, so option C is not appropriate. Tests of proportions use z procedures for categorical success/failure data, not a one-sample t-test for means. The t-test is part of inferential statistics because it uses sample evidence to make a decision about a population parameter. Study Guide references
/topics: one-sample t-test, sample mean, population mean, hypothesis testing.
NEW QUESTION # 96
A die is rolled twice. Probability of rolling two sixes?
Answer: A
Explanation:
Rolling a standard six-sided die twice creates two independent events. The probability of rolling a six on the first roll is 1/6 because there is one favorable outcome, six, out of six equally likely outcomes. The probability of rolling a six on the second roll is also 1/6. Because the outcome of the first roll does not affect the outcome of the second roll, the multiplication rule for independent events applies. Thus, P(six and six) = 1/6 × 1/6 = 1
/36. Another way to verify the result is to list the sample space: two die rolls produce 6 × 6 = 36 equally likely ordered outcomes. Only one outcome, (6, 6), satisfies the condition of two sixes. Therefore, the probability is
1 favorable outcome out of 36 total outcomes. Study Guide references/topics: independent events, multiplication rule, sample space, theoretical probability.
NEW QUESTION # 97
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 # 98
Probability of drawing king or queen from deck = ?
Answer: D
Explanation:
A standard deck contains 52 cards. There are 4 kings and 4 queens, one of each rank in each suit. The event is drawing a king or a queen. Since a single card cannot be both a king and a queen, these two events are mutually exclusive. Therefore, the number of favorable outcomes is 4 + 4 = 8. The probability is favorable outcomes divided by total outcomes: 8/52. This fraction can be simplified to 2/13, but the answer choice provides the unsimplified correct form. Option B, 4/52, counts only kings or only queens, not both. Option C,
1/13, is equivalent to 4/52 and again represents one rank only. Option D, 1/26, is too small and does not match the count of favorable cards. The word "or" signals that both qualifying ranks should be included. Study Guide references/topics: card probability, mutually exclusive events, addition rule, sample space.
NEW QUESTION # 99
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