Does the phone charge slower than the 90-minute spec?

Question

A phone maker specifies a mean full-charge time of 90 minutes. A reviewer runs 20 charge cycles and records a mean of 94 minutes with a sample standard deviation of 7.5 minutes. The population standard deviation is unknown.

Test at the 5% level (two-sided) whether the true mean charge time differs from 90 minutes, and report the t-statistic.

Accepted Answer

Unknown \sigma, so use a one-sample t-test with 20 - 1 = 19 degrees of freedom. Hypotheses: H0: \mu = 90 versus H1: \mu 
eq 90. Standard error: 	ext{SE} = \frac{7.5}{\sqrt{20}} = \frac{7.5}{4.472} = 1.677 	ext{ min}. Test statistic: t = \frac{94 - 90}{1.677} = \frac{4}{1.677} = 2.39. The two-sided 5% critical value for 19 degrees of freedom is t{0.025,\,19} = 2.093. Since 2.39 2.093, reject H0. The two-sided p-value is about 0.028. Conclusion: there is evidence the phone charges slower than its 90-minute specification, by about 4 minutes on average. Takeaway. Moving from 12 to 20 observati…

Does the phone charge slower than the 90-minute spec?

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