Level I Quant practice questions

When testing a hypothesis concerning a single population mean from a normally distributed population where the population variance is unknown, the t-statistic is the theoretically correct choice regardless of sample size.

Option A is wrong because a z-test requires either a known population variance or a very large sample size where the sample standard deviation can serve as a robust asymptotic proxy via the Central Limit Theorem.

Option C is wrong because the Chi-square statistic is utilized to test hypotheses regarding a single population variance, not a population mean.

Step 1: Identify the correct degrees of freedom for a simple linear regression coefficient test, which is $n - 2$. Given $n = 27$, $\text{df} = 27 - 2 = 25$. Therefore, the appropriate critical t-value is $t_c = 2.060$. Step 2: Calculate the margin of error by multiplying the critical value by the standard error of the slope:

Step 3: Calculate the lower bound of the confidence interval by subtracting the margin of error from the estimated slope:

Option A uses the standard normal z-distribution critical value of 1.96 instead of the t-distribution value: $2.10 - (1.96 \times 0.45) = 1.218 \approx 1.22$. Option B simply subtracts the unscaled standard error from the slope coefficient: $2.10 - 0.45 = 1.65$.

Step 1: Determine whether order matters. Because the portfolio is equally weighted, the order of stock selection is irrelevant, requiring the combinations counting rule ($_nC_r$). Step 2: Apply the combinations mathematical formula:

Step 3: Expand and simplify the factorials to find the numerical outcome:

Option A uses the permutation formula ($_nP_r$), incorrectly assuming that the order of stock selection changes the portfolio identity: $8! $/ 5! = 8 × 7 × 6 = 336$. Option C simply multiplies the pool size by the sub-portfolio size: $8 \times 3 = 24$.

Step 1: Recall the formula for the number of unique correlation coefficients in a multivariate normal distribution of $n$ variables:

Step 2: Substitute $n = 5$ into the formula:

Option B uses the number of assets, 5, which is the number of means (or variances), not of correlations. Option C counts only the correlations of one asset with the other four: $5 - 1 = 4$.

Step 1: Calculate the after-tax return for each asset class individually. Taxable Corporate Bonds: Interest is fully taxed, so the after-tax yield is:

Tax-exempt Municipal Bonds: No tax is levied, so the return remains:

Step 2: Use the asset allocation weights to find the total weighted portfolio after-tax return:

Step 3: Sum the individual weighted products together:

Option C calculates the standard pre-tax return of the portfolio, omitting taxes completely: $(0.60 \times 5\%) + (0.40 \times 4\%) = 4.60\%$. Option A computes the pre-tax portfolio return (4.60%) and incorrectly subjects the entire amount to the 30% tax rate, ignoring the tax-exempt status of the municipal allocation: $4.60\% \times (1 - 0.30) = 3.22\%$.

The Central Limit Theorem defines the mean of the sampling distribution of the sample mean to be exactly equal to the population mean $\mu$. The variance of this sampling distribution is equal to the population variance divided by the sample size, $\sigma^2 / n$.

Option B is wrong because $\sigma / \sqrt{n}$ represents the standard deviation of the sample mean (the standard error), not its variance.

Option C divides the mean by $n$ as well; the mean of the sampling distribution is $\mu$, and only the variance shrinks with $n$.

Step 1: Calculate the sample mean: $(4 + 7 + 8 + 9 + 12) / 5 = 40 / 5 = 8\%$. Step 2: Compute the squared deviations from the mean:

Step 3: Sum the squared deviations: $16 + 1 + 0 + 1 + 16 = 34$. Step 4: Divide by $n - 1$ because it is a sample variance:

Option A treats the sample as a population, incorrectly dividing the sum of squared deviations by $n$ instead of $n-1$: $34 / 5 = 6.80$. Option C calculates the sample standard deviation instead of the sample variance: $\sqrt{8.50} = 2.92$.

Data wrangling (or data preprocessing) involves transforming, scaling, or extracting data to make it suitable for modeling. Normalization and standardization (scaling data to have a mean of zero and standard deviation of one) are classic data wrangling steps.

Option B is incorrect because data cleansing specifically deals with handling missing values, removing invalid data, and correcting parsing errors.

Option C is incorrect because data exploration involves using summary statistics and visualizations to understand the fundamental properties of the dataset, rather than structurally transforming its scale.

Step 1: Value the perpetuity one period before the first cash flow. The first distribution is at Year 5, so the capitalized value is positioned at Year 4:

Step 2: Discount this amount back four years to today:

Rounding to the nearest dollar gives $660,078.

Option B is the undiscounted Year 4 perpetuity value. Option C compounds the Year 4 value forward instead of discounting it: $833,333.33 \times (1.06)^4 \approx 1,052,064$.

The standard decision rule states that if the p-value is less than or equal to the significance level ($\alpha$), the null hypothesis is rejected. Because $0.035 \le 0.05$, the result is statistically significant, and the null hypothesis must be rejected.

Option B reverses the decision rule: a p-value at or below the significance level means the null hypothesis is rejected, not kept.

Option C is wrong because a p-value does not represent the probability that the null hypothesis is true; rather, it is the probability of obtaining a test statistic at least as extreme as the one observed, assuming that the null hypothesis is completely true.