15 December 2026
Hypothesis testing asks whether sample evidence is difficult to reconcile with a stated population benchmark.
Choose direction from the question before seeing the result.
An analyst claims the population mean click-through rate is below 12 clicks per minute.
H_0:\mu=12,\qquad H_1:\mu<12.
The parameter is \mu, not the observed \bar{x}. The word below makes this a left-sided test.
Write hypotheses and identify the tail for each claim:
When \sigma is unknown:
t=\frac{\bar{x}-\mu_0}{s/\sqrt n},\qquad df=n-1.
The statistic is the observed difference measured in standard errors. Its sign indicates direction; its magnitude indicates distance from H_0.
| Alternative | Excel p-value |
|---|---|
| H_1:\mu<\mu_0 | T.DIST(t,df,TRUE) |
| H_1:\mu>\mu_0 | 1-T.DIST(t,df,TRUE) |
| H_1:\mu\ne\mu_0 | T.DIST.2T(ABS(t),df) |
Using ABS with T.DIST.2T avoids separate positive/negative formulas.
The p-value is the probability—assuming H_0 is true—of obtaining a test statistic at least as extreme as the one observed, in the direction specified by H_1.
It is not:
Given n=45, \bar{x}=10.8, s=3.2, \mu_0=12, and \alpha=0.05:
=T.DIST(t,44,TRUE);A random sample of 40 students has \bar{x}=283 and s=17.85. Test at \alpha=0.05 whether population mean monthly discretionary spending exceeds AUD 275.
Complete all five steps, show the Excel p-value formula, and state the conclusion without saying “accept H_0”.
A two-sided test considers evidence in both directions and uses both tails. A one-sided test concentrates the rejection region in one pre-specified direction.
Do not choose one-sided after observing the sample. That inflates the chance of finding a convenient result.
Using the same spending sample, test whether the mean differs from AUD 275.
For H_0:\pi=\pi_0:
z=\frac{\hat p-\pi_0}{\sqrt{\pi_0(1-\pi_0)/n}}.
Check:
n\pi_0\ge5,\qquad n(1-\pi_0)\ge5.
The null value \pi_0 appears in the test standard error.
| Alternative | Excel p-value |
|---|---|
| H_1:\pi<\pi_0 | NORM.S.DIST(z,TRUE) |
| H_1:\pi>\pi_0 | 1-NORM.S.DIST(z,TRUE) |
| H_1:\pi\ne\pi_0 | 2*(1-NORM.S.DIST(ABS(z),TRUE)) |
The tail follows the alternative hypothesis, not merely the sign of z.
A bank benchmarks error-free processing at 95%. In a random sample of 250 transactions, 229 succeed.
A company claims at least 98% of invoices are accurate. A random audit finds 288 accurate invoices out of 300.
Test at \alpha=0.05 whether the population accuracy rate is below 98%.
Include the assumption check, statistic, Excel p-value, decision, and business conclusion.
For a two-sided test at significance \alpha:
The interval also communicates plausible effect sizes, which the p-value alone does not.
For the invoice sample:
If p-value <\alpha:
Reject H_0; there is sufficient evidence that …
If p-value \ge\alpha:
Fail to reject H_0; there is insufficient evidence that …
“Fail to reject” is not proof that the null is true.
| Decision | H_0 true | H_0 false |
|---|---|---|
| Reject H_0 | Type I error | correct |
| Fail to reject H_0 | correct | Type II error |
For H_1:\mu>275:
Name the real-world cost of each error before choosing \alpha.
For the invoice-accuracy test:
A report says, “Waiting time fell by 0.2 minutes, p<0.001, so the redesign was a major success.”
Identify:
Check that the spreadsheet separates:
Trace formulas to ensure the sample statistic, benchmark, and tail are consistent.
Choose either a mean claim from week4-testing-data.xlsx or a product-trial proportion from week4-proportions-data.xlsx.
Produce a one-page brief containing the five test steps, an interval or effect estimate, one possible error, and a two-sentence recommendation.
Can you: