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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.

KS Comments

This question is fairly manageable.

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