Numerical answers follow the slide scenarios. The optional integrated brief below uses the product-trial sample in the Week 4 proportions workbook; other well-specified claims from the testing workbook can also earn full credit.
Download the Week 4 worked Excel workbook.
Claims and mean tests
Watch me: translate a click-through claim / Now you try: write the hypotheses
The demonstration is \(H_0:\mu=12\) versus \(H_1:\mu<12\), a left-sided test. For the exercise: (1) \(H_0:\mu=30\), \(H_1:\mu>30\), right; (2) \(H_0:\mu=85\), \(H_1:\mu\ne85\), two-sided; (3) \(H_0:\pi=0.04\), \(H_1:\pi<0.04\), left; (4) \(H_0:\pi=0.50\), \(H_1:\pi>0.50\), right. Boundary wording such as “at least 98%” can be represented by a composite null \(\pi\ge0.98\); calculations use its boundary value.
Watch me: test click-through rate / Now you try: discretionary spending
Click-through: \(t=(10.8-12)/(3.2/\sqrt{45})=\mathbf{-2.516}\), \(df=44\), and =T.DIST(-2.515576,44,TRUE) gives 0.00780. At 5%, reject \(H_0\); the sample supports a mean below 12 clicks per minute, assuming the test conditions hold.
Spending: \(H_0:\mu=275\), \(H_1:\mu>275\); \(t=(283-275)/(17.85/\sqrt{40})=\mathbf{2.835}\), \(df=39\). =1-T.DIST(2.834535,39,TRUE) gives 0.00362. Reject at 5%; there is evidence that population mean monthly discretionary spending exceeds AUD 275. The estimated difference is AUD 8. A random sample is stated; the test also needs a sufficiently reasonable sampling distribution of the mean.
Now you try: change only the claim
For \(H_1:\mu\ne275\), the statistic stays 2.835 but use =T.DIST.2T(ABS(2.834535),39) to get 0.00723. The alternative, tail calculation and conclusion wording change. Both one- and two-sided tests reject at 5%; the latter supports “differs from AUD 275” without choosing direction in advance.
One-sample decision threshold
Watch me: property-price threshold / Now you try: a different property group
For all 329 properties, the sample mean price is 1,215.462 and sample SD is 418.922 in AUD thousands. Test \(H_0:\mu\le1200\) against \(H_1:\mu>1200\). The standard error is 23.0959, giving \(t=(1215.462-1200)/23.0959=\mathbf{0.6695}\) with \(df=328\). Excel’s =T.DIST(-0.669469,328,TRUE) gives 0.2518. At 5%, fail to reject: the sample does not give sufficient evidence that population mean price exceeds AUD 1.2 million. The sample mean alone being above the threshold is insufficient.
For the 67 properties with SchoolZone=No, the sample mean is 1,046.045 and SD is 361.761 in AUD thousands. Test \(H_0:\mu\le1000\) against \(H_1:\mu>1000\). The standard error is 44.1962, \(t=\mathbf{1.0418}\), \(df=66\), and =T.DIST(-1.041827,66,TRUE) gives 0.1506. Fail to reject at 5%. The inference applies only to a population that the selected properties could reasonably represent; filtering records does not itself create a random sample.
Proportion tests and interval
Watch me: transaction success rate / Now you try: invoice accuracy
Bank: \(\hat p=229/250=0.916\). Under \(H_0:\pi=0.95\), expected successes and failures are 237.5 and 12.5, both at least 5. \(z=\mathbf{-2.467}\); =NORM.S.DIST(-2.466619,TRUE) gives 0.00682. Reject at 10%; there is evidence of a success rate below 95%.
Invoices: \(H_0:\pi=0.98\) (or \(\pi\ge0.98\)), \(H_1:\pi<0.98\). The audit gives \(\hat p=288/300=0.96\). Null expected successes/failures are 294 and 6; the slide’s threshold of 5 is met, but the smaller failure count makes the Normal approximation worth noting. \(z=(0.96-0.98)/\sqrt{0.98(0.02)/300}=\mathbf{-2.474}\); =NORM.S.DIST(-2.474358,TRUE) gives 0.00667. Reject at 5%; the data support an accuracy rate below 98% under the stated approximation.
Now you try: interval cross-check
The ordinary 95% Wald interval uses the sample rate in its standard error: \(0.96\pm1.96\sqrt{0.96(0.04)/300}=\mathbf{[93.78\%,98.22\%]}\). It includes 98%. This is consistent with a two-sided 5% test failing to reject, while the prespecified left-sided test rejects. The interval shows plausible effect sizes and uncertainty; do not claim it contradicts the one-sided result.
Decisions, errors and importance
Watch me: errors in context / Now you try: business consequences
For the spending claim, a Type I error means concluding mean spending exceeds AUD 275 when it does not; a Type II error means missing a true excess. For invoices, a Type I error is concluding accuracy is below 98% when it is at least 98%, potentially causing needless process changes and cost. A Type II error is failing to detect accuracy below 98%, leaving customers and staff exposed to errors. 5% is a defensible default; 10% may be chosen if missed errors are especially costly, while 1% reduces false alarms at the cost of less detection power. Choose \(\alpha\) before inspecting the sample and explain the relative costs; there is no universal correct level.
Now you try: challenge the headline
\(p<0.001\) is evidence against a zero mean reduction under the test model. It does not establish a major improvement or that the redesign caused it. Report the 0.2-minute = 12-second estimate, a confidence interval, a pre-set meaningful threshold, baseline wait and implementation cost, and the study design. A defensible statement is: “The observed mean wait was 12 seconds lower with strong statistical evidence, but its operational value and causal attribution require further assessment.”
Integrated evidence brief
Watch me: audit a testing worksheet / Now you try: complete the evidence brief
A worked one-page option is to test whether more than half the target market has tried the product. From Marketing, 495 of 856 report trial, \(\hat p=0.57827\). State \(H_0:\pi=0.50\), \(H_1:\pi>0.50\), \(\alpha=0.05\). Null expected counts are 428 successes and 428 failures. \(z=(0.57827-0.50)/\sqrt{0.5(0.5)/856}=\mathbf{4.58}\); =1-NORM.S.DIST(4.58,TRUE) gives about 0.0000023. Reject \(H_0\); there is evidence that the population proportion is above half, conditional on representative sampling. The 95% interval is [54.52%, 61.14%]. A Type I error would be declaring majority trial when true trial is at most half. Recommendation: “Plan for a majority-tried segment, but validate the survey frame before applying this estimate to the full market. Track non-response and repeat with a probability sample if a high-cost launch decision depends on the precise proportion.”
Retrieval and closing checks
A confidence interval estimates a population parameter with a method having a stated long-run coverage rate. A standard error measures sample-statistic variability across repeated samples. For a mean with unknown population SD, use a \(t\) method with appropriate degrees of freedom and assumptions. Hypotheses concern population quantities (\(\mu\), \(\pi\)), because sample quantities (\(\bar x\), \(\hat p\)) are observed evidence. Choose the test tail from the claim before computing, and use “reject” or “fail to reject” with a contextual conclusion. A small p-value measures evidence under the null, not the size of a business benefit.