Pool battery samples before calculating a test

For $n=70$, $\bar x=10317/70=147.3857142857$ and $\begin{aligned}&s^2\\ & =\frac{1540231-10317^2/70}{69}\\ & =\frac{1375681}{4830}\\ & =284.8200828157.\end{aligned}$ Under $H_0$, use $\bar X\approx N(150,s^2/70)$ by CLT with variance estimated from the sample. Then $z=-1.296035106$ and the lower-tail p-value is 0.09748169520, about 0.0975 to 3 s.f. The printed .0978 comes from rounding the mean to 147.39 before the test; its screen shows $\sigma=16.876611034209$, $z\approx-1.293910646$, $p\approx0.0978482084$. Both versions are retained as distinct numerical calculations.

Assuming the population mean is 150 h and the stated large-sample reference model is suitable, the estimated probability of a random size-70 sample mean at or below the observed 147.3857142857 h is about .09748. The printed explanation uses its rounded threshold 147.39 and value .0978. Neither is the probability that the null hypothesis itself is true.

Assume independent samples from the same unchanged population. Pool $n=70+50=120$, $\sum x=10317+7331=17648$, $\sum x^2=1540231+1100565=2640796$. Thus $\bar x=17648/120=147.0666666667$ and $\begin{aligned}&s^2\\ & =\frac{2640796-17648^2/120}{119}\\ & =\frac{680452}{1785}\\ & =381.2056022409.\end{aligned}$ Do not simply average the two sample variances or means without the sample-size weights.

Under $H_0$, $\bar X\approx N(150,381.2056022409/120)$. Then $z=-1.645782360$ and the lower-tail $p=0.04990428762\le0.10$, so reject: sufficient evidence at 10% that population mean life is below 150 h. Equivalently, $-1.64578<-1.281551566$, the 10% lower critical value. The source enters rounded $\bar x=147.07$ and variance 381.21 ($\sigma=19.524599867859$), yielding $z\approx-1.643902671$, $p\approx0.0500981388\to0.0501$. The requested 10% decision agrees; numerically the two calculations fall on opposite sides of 5%, illustrating why intermediate precision matters. This does not remove uncertainty in the approximate model.