The numerical trial answers use Week 3 proportions data. The audit calculations use the values supplied on the slides. Intervals below are ordinary 95% Wald intervals unless a different level is stated.
Download the Week 3 worked Excel workbook.
Sampling and variation
Watch me: diagnose a survey / Now you try: redesign the evidence
The app demonstration targets all relevant users, but the frame contains only active users who see the prompt. Inactive or departed users are missed, and “excellent” leads respondents toward a positive answer. Use neutral wording, a defined recent-user population, a sample drawn from the customer list, and follow-up for non-response. In the smartwatch exercise, selecting only customers who exercised yesterday excludes less active users; voluntary email response adds self-selection. A huge sample from that biased frame reduces random error but not selection or wording bias. Sample randomly from all purchasers in a defined period, measure baseline and later activity consistently, and compare with an appropriate control or before/after design. Generalise only to the population represented by the sampling frame, subject to non-response.
Watch me: compare sample sizes / Now you try: sampling variation
With \(\sigma=20\), \(SE(\bar X)\) is 4 at \(n=25\) and 2 at \(n=100\). With \(\sigma=48\), it is 8 at \(n=36\) and 4 at \(n=144\). The larger sample gives a more precise mean estimate. Halving standard error requires four times the sample size because \(SE=\sigma/\sqrt n\). Neither calculation fixes non-response bias.
Mean intervals
Watch me: audit balances, known \(\sigma\) / Now you try: audit balances, unknown \(\sigma\)
For known \(\sigma=22.38\), \(z^*=\) NORM.S.INV(0.975) \(=1.9600\), \(SE=22.38/\sqrt{80}=2.5022\), margin AUD 4.9041, and interval AUD [81.15, 90.95]. For unknown \(\sigma\) with \(s=24.02\), use \(df=79\), =T.INV.2T(0.05,79) \(=1.9905\), \(SE=24.02/\sqrt{80}=2.6851\), margin AUD 5.3454, interval AUD [80.70, 91.40]. State: using this sampling procedure, a 95% interval for the population mean audit balance is approximately AUD 80.70–91.40. The sample size alone cannot validate random sampling.
Now you try: confidence versus precision
With \(n=80\), \(\bar x=86.05\) and \(s=24.02\), the 90% \(t\) interval is [81.58, 90.52], 95% is [80.70, 91.40], and 99% is [78.96, 93.14], all in AUD. Higher confidence gives a wider interval at fixed data. Smaller \(n\) generally widens it. Selection bias and measurement error are not reflected in these margins.
Proportion intervals
Watch me: recruitment-service use / Now you try: product trial proportion
For 126 of 600 employers, \(\hat p=0.21\), successes 126, failures 474, \(SE=\sqrt{0.21(0.79)/600}\), and the 95% interval is [17.74%, 24.26%]. In Marketing, 495 of 856 have tried the product, so \(\hat p=\mathbf{57.83\%}\), with 495 successes and 361 failures. The 95% interval is [54.52%, 61.14%]. The entire interval is above 50%, providing evidence that the population trial proportion exceeds half if the sample represents the target market. The workbook alone does not establish random selection or coverage of the market.
Now you try: subgroup intervals
| Female |
216 / 398 |
54.27% |
[49.38%, 59.17%] |
| Male |
279 / 458 |
60.92% |
[56.45%, 65.39%] |
Both satisfy the large-count check. The male sample rate is higher by about 6.65 percentage points. The intervals overlap, so do not claim a statistically established difference solely from this visual comparison; a direct difference interval or test is needed. Neither difference establishes causation.
Reusable template and scenarios
Watch me: build an Excel template
Put sample inputs in labelled cells: estimate, sample size, confidence level, and either \(s\) for a mean or success count for a proportion. For an unknown-\(\sigma\) mean, use =T.INV.2T(1-confidence,n-1) and =s/SQRT(n). For a proportion, use =NORM.S.INV((1+confidence)/2) and =SQRT(phat*(1-phat)/n). Margin is critical value times standard error; bounds are estimate minus/plus margin. Check random sampling and, for the simple proportion interval, sufficiently many successes and failures.
Now you try: scenario analysis
At fixed estimate and sample size, 90% gives the narrowest interval, then 95%, then 99%. At fixed variability, increasing \(n\) from 50 to 200 to 800 halves the standard error each time. For a mean, \(SE=s/\sqrt n\) (or \(\sigma/\sqrt n\) when known); for a proportion, \(SE=\sqrt{\hat p(1-\hat p)/n}\). A manager can plan for wider intervals at higher confidence and can quadruple a random sample to roughly halve margin of error; neither tactic repairs biased sampling.
Retrieval and closing checks
A parameter describes the population; a statistic is calculated from a sample. In \(P(A\mid B)\), the denominator is the number or probability in group \(B\). A \(t\) distribution accounts for uncertainty from estimating the population SD in mean inference. Large datasets can still miss part of the target population. Sampling variation shrinks as \(n\) grows; selection bias does not. The law of large numbers says averages stabilise toward a population mean under appropriate sampling, while the central limit theorem supports an approximately Normal sampling distribution of the mean under suitable conditions as \(n\) grows. Report interval bounds as evidence with assumptions, not certainty about a fixed parameter.