A sample of test scores has a mean of 74 and a median of 82. Describe the probable shape of the distribution, explaining the relationship between the mean and median.

Because the mean (74) is substantially less than the median (82), the mean has been pulled downward by extreme low scores in the left tail. Therefore, the distribution is skewed to the left.

A distribution of salaries has $Q_1 = \$42,000$ and $Q_3 = \$68,000$. A director earns $\$110,000$. Is this salary an outlier according to the 1.5 IQR rule?

Compute IQR: $$\text{IQR} = 68000 - 42000 = 26000.$$

Upper outlier boundary: $$Q_3 + 1.5(\text{IQR}) = 68000 + 1.5(26000) = 68000 + 39000 = \$107,000.$$

Because $\$110,000 > \$107,000$, this salary is an outlier.

A regression equation predicting used car price ($y$, in thousands of dollars) from age ($x$, in years) is $\hat{y} = 28.5 - 2.1x$. Interpret the slope $-2.1$ in context.

Interpretation: For each additional year in the car's age, the predicted price decreases by $2,100 (or 2.1 thousand dollars) on average.

A high school has 400 freshmen, 350 sophomores, 320 juniors, and 300 seniors. An administrator randomly selects 25 students from each grade level to survey opinion on a school schedule change. Identify the sampling design and explain why it is advantageous.

This is a stratified random sample. The grades represent homogeneous strata with respect to high school experience. Stratifying guarantees equal representation from each grade level and reduces sampling variability compared to a simple random sample.

A researcher wants to evaluate a new memory supplement. 60 volunteers are paired based on baseline memory test scores (the two highest scores paired, the next two highest paired, etc.). Within each pair, a coin flip assigns one subject to the supplement and the other to an identical placebo. What is this experimental design called, and what is its primary benefit?

This is a matched-pairs design (a specialized form of randomized block design). Its primary benefit is that it controls for baseline memory differences among participants, drastically reducing experimental variability and increasing the power to detect any true effect of the supplement.

An observational study finds that people who drink green tea regularly have a lower incidence of cardiovascular disease than non-tea drinkers. Can the researchers conclude that green tea consumption causes a reduction in heart disease risk? Identify a possible confounding variable.

No, causal conclusions cannot be established from an observational study because treatments were not randomly assigned. A plausible confounding variable is overall lifestyle / diet: regular green tea drinkers may exercise more frequently and eat healthier diets than non-tea drinkers, which independently lowers cardiovascular disease risk.

In a school, 60% of students take art ($A$), 40% take music ($M$), and 20% take both art and music. What is the probability that a randomly chosen student takes art, given that they take music? Are taking art and taking music independent events?

Compute conditional probability: $$P(A \mid M) = \frac{P(A \cap M)}{P(M)} = \frac{0.20}{0.40} = 0.50 = 50\%.$$

Test independence: Since $P(A \mid M) = 0.50 \neq P(A) = 0.60$, taking art and taking music are not independent.

A company produces components with weight $X$ ($\mu_X = 50\text{ g}, \sigma_X = 4\text{ g}$) and cases with weight $Y$ ($\mu_Y = 20\text{ g}, \sigma_Y = 3\text{ g}$). The combined assembly weight is $W = X + Y$. If $X$ and $Y$ are independent, find the mean and standard deviation of $W$.

Mean: $$\mu_W = \mu_X + \mu_Y = 50 + 20 = 70\text{ g}.$$

Variance: Since independent, variances add: $$\sigma_W^2 = \sigma_X^2 + \sigma_Y^2 = 4^2 + 3^2 = 16 + 9 = 25.$$

Standard deviation: $$\sigma_W = \sqrt{25} = 5\text{ g}.$$

A fair coin is flipped 10 times. What is the probability of getting exactly 7 heads?

This follows a binomial distribution with $n = 10$, $p = 0.5$, and $k = 7$: $$P(X = 7) = \binom{10}{7} (0.5)^7 (0.5)^3 = \frac{10!}{7!3!} (0.5)^{10} = 120 \times \frac{1}{1024} = \frac{15}{128} \approx \mathbf{0.1172}.$$

A 95% confidence interval for the proportion of registered voters who support an education bond is $(0.52, 0.58)$. Interpret this interval in context, and determine whether there is convincing evidence that a majority supports the bond.

Interpretation: We are 95% confident that the interval from 0.52 to 0.58 (or 52% to 58%) captures the true proportion of all registered voters who support the bond.

Conclusion: Yes, because the entire 95% confidence interval lies strictly above 0.50 (50%), there is convincing statistical evidence that a majority of registered voters support the bond.

A factory claims its candy machines produce bags with an average weight of 50.0 grams. A consumer group tests $H_0: \mu = 50.0$ vs $H_a: \mu < 50.0$ using a sample of 25 bags, finding $\bar{x} = 49.2\text{ g}$ and $s = 1.5\text{ g}$. Compute the $t$ test statistic and conclude at $\alpha = 0.05$.

Compute test statistic: $$t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} = \frac{49.2 - 50.0}{1.5 / \sqrt{25}} = \frac{-0.8}{0.3} = -2.67 \quad (df = 24).$$

From the $t$-distribution with $df = 24$, $P(T < -2.67) \approx 0.0067$.

Because the p-value $0.0067 < \alpha = 0.05$, we reject $H_0$. There is convincing evidence that the true mean bag weight is less than 50.0 grams.

A computer printout for a linear regression model yields: Predictor: Constant Coef = 12.4, SE = 2.1; Slope Coef = 3.65, SE = 0.73, t = 5.00, p = 0.0001 with $n = 20$. Test $H_0: \beta = 0$ vs $H_a: \beta \neq 0$ at $\alpha = 0.01$.

The test statistic for the slope is: $$t = \frac{b - 0}{\text{SE}_b} = \frac{3.65}{0.73} = 5.00 \quad (df = n - 2 = 18).$$

Because the p-value $0.0001 < 0.01$, we reject $H_0$. There is convincing statistical evidence of a linear relationship between the explanatory and response variables.