Answer
There is a sufficient evidence that the mean is greater than 100.
Work Step by Step
Here we have $H_{o}$: μ = 100, $H_{1}$: μ > 100, n = 35, x̅ = 104.3, s = 12.4 and α = 0.05
Using the Classical approach:
Using Table VI, we have: t = 1.691
$σ_{ x̅} = \frac{s}{\sqrt n} = \frac{12.4}{\sqrt 35} = 2.1$
$t = \frac{104.3 - 100}{2.1} = 2.05$
$P(t \gt 2.05) \approx 0.024$
Since, $2.05 \gt 1.691$, and $P\lt 0.05$, we will reject the null hypothesis.
