Separate a large-sample test from a known-variance test

Choose the 120 patients by chance from the relevant patient population, giving each eligible patient an equal chance of inclusion rather than selecting conveniently available patients. In a simple random sample without replacement, every group of 120 has the same chance of selection. The source’s phrase “equal and independent chance” describes its intended random selection; without-replacement inclusion events are not literally independent.

$\bar x=7+24/120=7.2$ days. The unbiased variance estimate is $\begin{aligned}&s^2\\ & =\frac{14908-24^2/120}{119}\\ & =125.2369748\approx 125\ \mathrm{days}^2.\end{aligned}$

Let $\mu$ be the population mean recovery period in days. Test $H_0:\mu=7$ against $H_1:\mu>7$. Under $H_0$, for n 120 use $\bar X\approx N(7,125.2369748/120)$ by the CLT. Then $z=0.1957736936$ and upper-tail $p=0.4223936506>0.05$. Do not reject: insufficient evidence at 5% that 7 days understates the mean. Equivalently $z<1.644853627$, so the critical-value method gives the same conclusion. Source p.422393669 agrees to calculator precision.

Not on the information given. With only 15 patients and no normal population assumption, the source’s large-sample normal approximation is not justified. This does not rule out every possible small-sample method; it means this test needs additional assumptions or a suitable alternative.

For the new n 120 sample, test $H_0:\mu=7$ against $H_1:\mu\ne 7$ with known $\sigma=1$ day. Under $H_0$, use $\bar X\approx N(7,1/120)$. Rejection at 1% requires $\left|\frac{\bar x-7}{1/\sqrt{120}}\right|\ge 2.575829304.$ Therefore $0\le\bar x\le 6.764860031$ or $\bar x\ge 7.235139969$ days. The lower bound 0 comes from nonnegative recovery periods. Source 6.76/7.24 are rounded boundary labels, not exact replacements for the interval endpoints.