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Unbiased estimate of population mean = \bar{x} = 41.3

Unbiased estimate of population variance = s^2 = 1.584

Let \mu be the population mean time required by an employee to complete a task.
H_0: \mu = 42 against

H_1: \mu \neq 42 at 10% level of significance

Reject H_0 if p-value \textless 0.10

Test Statistic, T = \frac{\bar{x}-\mu_0}{\frac{s}{\sqrt{n}}} \sim t(10)
Applying t-test with \bar{t} = 41.3, n = 11, s^2 = 1.584

From GC, p-value = 0.09487 \textless 0.10, reject H_0.

Thus, there is sufficient evidence at 10% significance level that there has been a change in the mean time required by an employee to complete the task.

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