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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.