Examworthy
Open the appSign inStart free

NCA-ADS domain - 13% of the exam

Descriptive Analysis and Visualization

Descriptive Analysis and Visualization is 13% of the NCA-ADS exam. These are the objectives it covers, each with practice questions, with every answer explained.

The domain in numbers

41
Practice questions
4
Objectives
20
Easy
21
Medium

Where people go wrong

  • Reaching for the mean when the data is skewed

    The wrong answers pick the mean because it uses every value, or pick a spread measure such as variance, range or standard deviation as if it described the centre. In a distribution with a few extreme high values, the mean is pulled upward and overstates the typical figure. Variance, range and standard deviation measure dispersion, not central tendency. The median depends only on the middle position of the ordered values, so a handful of extreme earners barely moves it.

    See it in a sample question

  • Reading a p-value as the probability of truth or size

    The wrong answers treat a p-value as the probability that the null hypothesis is true, as proof of equality when the null is not rejected, or as a measure of how much a difference matters. A p-value is the probability of data at least as extreme as the sample, assuming the null is true. Failing to reject never proves the null, an interval containing zero leaves a zero difference plausible, and a very large sample can make a tiny effect significant, so importance needs effect size.

    See it in a sample question

  • Treating a lower significance level as free safety

    The wrong answers treat tightening alpha as an improvement with no cost. Alpha is the Type I error rate, so lowering it reduces false positives. With the sample size fixed, the rejection region shrinks and a real effect becomes harder to detect, so the Type II error rate rises and power falls. To lower the chance of missing a real effect, the correct answer increases the sample size, which raises power without touching the significance level.

    See it in a sample question

  • Choosing a test that ignores the data structure

    The wrong answers choose a test from the number of means or categories mentioned and skip the structure of the data. A t-test needs a continuous outcome, and an independent t-test ignores matched measurements on the same items, where a paired t-test is needed. A paired t-test cannot handle several independent groups, where one-way ANOVA applies. A chi-square test of independence needs two categorical variables, while goodness-of-fit checks one categorical variable against claimed proportions.

    See it in a sample question

  • Swapping the histogram and the bar chart

    The wrong answers swap the two charts that show counts or magnitudes. A histogram bins a single continuous numeric variable, such as response times, to show its distribution. A bar chart gives each named category its own bar, so it suits comparing nominal categories such as product lines or payment methods. A histogram has nothing to bin for categories, and a bar chart cannot show how a continuous variable is spread across a range.

    See it in a sample question

Objectives in this domain

Sample questions from this domain

Free sampleDescriptive Analysis and Visualizationeasy

A data scientist is summarising the annual salaries in a dataset that contains a small number of extremely high earner outliers. Which measure of central tendency gives the most representative picture of the typical employee salary?

  • AThe median, because it is resistant to extreme values at the tail of the distribution Correct
  • BThe arithmetic mean, because it incorporates every value in the dataset
  • CThe mode, because the most frequently occurring salary describes what is normal
  • DThe variance, because it quantifies how spread out the salary values are
Explain when the median is preferable to the mean as a measure of central tendency for skewed or outlier-affected data. When a distribution is right-skewed by outliers, the arithmetic mean is dragged toward the tail and overstates where the bulk of observations lie. The median, defined as the middle value of the ordered dataset, is unaffected by the magnitude of extreme values and therefore provides a more representative centre for salary-type data.

Why A is correct: The median splits the ordered distribution at the midpoint and is unaffected by how extreme the highest values are, so it reflects where most salaries actually fall.

Why B is wrong: The mean is tempting because it uses all data points, but outliers pull it far above most employees' salaries, making it unrepresentative of the typical worker.

Why C is wrong: The mode identifies the most common single value, but salary data is often spread across many distinct figures, making the mode a poor summary of the centre.

Why D is wrong: Variance measures dispersion, not central tendency, so it describes spread rather than where a typical salary sits in the distribution.

Free sampleDescriptive Analysis and Visualizationmedium

A data scientist runs a two-sample t-test comparing mean GPU kernel execution times between two hardware configurations. The result is p = 0.03 with a significance level of 0.05. Which statement is the most statistically precise interpretation of this result?

  • AThere is a 3% probability that the null hypothesis is true
  • BThe alternative hypothesis is true with 97% confidence, confirming a real difference exists
  • CAssuming the null hypothesis is true, there is a 3% probability of obtaining a test statistic at least as extreme as the one observed Correct
  • DThe probability of committing a type II error on this test is 3%
Identify the correct frequentist interpretation of a p-value and distinguish it from common misinterpretations. A p-value answers the question: if the null hypothesis were true, how often would random sampling alone produce a result this extreme? A p of 0.03 means that outcome arises 3% of the time under the null, which is below the 0.05 threshold, so we reject the null. The p-value says nothing about the probability that either hypothesis is true, the size of the effect, or the type II error rate.

Why A is wrong: This is one of the most common misinterpretations of a p-value. A p-value is not the probability that the null hypothesis is true; it is the probability of observing data at least as extreme as the sample data, assuming the null hypothesis is already true.

Why B is wrong: A p-value below the significance threshold justifies rejecting the null hypothesis but does not assign a probability of truth to the alternative hypothesis. Confidence intervals and effect sizes are needed to gauge practical significance.

Why C is correct: This is the precise frequentist definition of a p-value. It quantifies how compatible the observed data are with the null hypothesis, not the probability that either hypothesis is correct.

Why D is wrong: Type II error probability, known as beta, depends on statistical power and the effect size being tested. It is a separate quantity from the p-value and cannot be read directly from a single test result.

Free sampleDescriptive Analysis and Visualizationmedium

An analyst lowers the significance level of a hypothesis test from 0.05 to 0.01 while keeping the sample size and the true effect size unchanged. Considering only the direct effect of this change on the two types of error, what happens?

  • ABoth the probability of a Type I error and the probability of a Type II error decrease.
  • BThe probability of a Type I error decreases and the probability of a Type II error increases. Correct
  • CThe probability of a Type I error increases and the probability of a Type II error decreases.
  • DNeither error probability changes, because the significance level affects only the width of the reported confidence interval.
Lowering the significance level reduces Type I error but increases Type II error when sample size is fixed. Alpha is the probability of a Type I error, so reducing it directly lowers false positives. Because statistical power and Type II error depend on the rejection region, a smaller alpha with an unchanged sample size shrinks that region and increases the chance of missing a true effect.

Why A is wrong: Tempting because tightening the threshold feels universally safer, but it is wrong: the two error probabilities trade off, so reducing one with a fixed sample size raises the other.

Why B is correct: Correct: alpha is the Type I error rate, so lowering it reduces false positives, but with the sample size fixed the rejection region shrinks, making a true effect harder to detect and raising the Type II error rate.

Why C is wrong: Tempting if a candidate confuses the direction of the change, but lowering alpha shrinks the rejection region, which decreases rather than increases the Type I error rate.

Why D is wrong: Tempting because alpha does relate to confidence intervals, but it is wrong: alpha is by definition the Type I error rate, so changing it directly changes that error probability.

Free sampleDescriptive Analysis and Visualizationmedium

A data scientist benchmarks one GPU kernel by timing it on the same 50 input matrices twice, once under an old driver and once under a new driver, so every matrix is measured under both conditions. She wants to test whether the new driver changes the mean execution time. Which hypothesis test correctly accounts for the structure of this data?

  • AAn independent two-sample t-test, because there are two driver conditions and the goal is to compare two group means.
  • BA one-way analysis of variance, because the readings come from repeated timing runs across many matrices.
  • CA chi-squared test of independence, because the experiment applies two driver versions to the same matrices.
  • DA paired (dependent) two-sample t-test, because each matrix is measured under both drivers and the two readings form linked observations. Correct
When the same units are measured under two conditions, a paired t-test analyses the within-unit differences and is the correct test. Because each matrix is timed under both drivers, the two measurements share the matrix-specific baseline; a paired t-test works on the per-pair differences, cancelling that shared variation and giving a more powerful test than treating the groups as independent.

Why A is wrong: Tempting because there are two conditions and two means, but it is wrong: the independent t-test assumes unrelated samples, whereas each matrix appears in both groups, so ignoring the pairing inflates the error variance and weakens the test.

Why B is wrong: Tempting because ANOVA also compares means, but it is wrong: ANOVA is for three or more groups, while there are only two conditions here, and a standard one-way ANOVA does not model the matched-pair structure either.

Why C is wrong: Tempting because two conditions are involved, but it is wrong: the chi-squared test of independence works on counts in a contingency table of categorical variables, whereas execution time is a continuous measurement, so this test does not apply.

Why D is correct: Correct. The same 50 matrices are timed under both drivers, so each old and new reading forms a matched pair; a paired t-test analyses the per-matrix differences and removes between-matrix variability, which is exactly the structure described.

Free sampleDescriptive Analysis and Visualizationeasy

An analyst summarises the revenue contributed by each of four product lines for the most recent quarter and wants a chart that compares the exact revenue of the four lines against one another. The four values are on the same scale and there is no time dimension. Which chart type communicates this comparison most clearly?

  • AA histogram of the four revenue figures, binning them so that the frequency of each revenue level is displayed on the axis.
  • BA line chart connecting the four product lines along the horizontal axis in label order so the differences are joined up.
  • CA scatter plot placing each product line at a point whose vertical position is its revenue and horizontal position is arbitrary.
  • DA bar chart, drawing one bar per product line whose height equals that product line's revenue for the recent quarter. Correct
A bar chart is the appropriate visualisation for comparing a single numeric value across a small set of discrete categories. Comparing magnitudes across named categories is a length-comparison task, which the human eye performs accurately along a common baseline. A bar chart provides exactly that baseline and one bar per category, whereas line, scatter, or histogram encodings either imply false ordering or hide the categories.

Why A is wrong: A histogram shows the distribution of many values across bins and is tempting because revenue is numeric, but with only four named categories there is nothing to bin and the category labels would be lost.

Why B is wrong: A line chart implies a continuous ordered axis and is tempting because it shows magnitude on the vertical axis, but joining unordered product categories with a line falsely suggests progression between them.

Why C is wrong: A scatter plot relates two continuous variables and is tempting because it encodes the numeric revenue vertically, but there is no second numeric variable, so the horizontal placement is meaningless and comparison is harder than with bars.

Why D is correct: A bar chart maps a categorical variable to position and a numeric value to bar length, so comparing the magnitude of a few discrete categories is precise and immediate; this is the standard choice for comparing values across categories.

Other domains in this exam

See also the NCA-ADS cert hub, the study guide, and the cheat sheet.

Examworthy is not affiliated with or endorsed by NVIDIA. Original, blueprint-aligned practice material only.