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Unbiased estimate of mean $= \frac{305}{250} + 75 = 76.22$

Unbiased estimate of variance $= \frac{1}{249} (29555 - \frac{306^{2}}{250}) = 117$

$H_0: \mu = 75$

$H_1: \mu > 75$

Under $H_0$, $\bar{X} ~\sim~ \mathrm{N} (75 \frac{117.2004}{250})$.

From graphing calculator, p-value $= 0.0374 > 0.025$, we do not reject $H_0$.

Thus, there is insufficient evidence at 2.5% level of significance that this type of device can retain information for more than 75 hours.