AHL 4.16—Confidence intervals

Since $\sigma$ is known, use the $z$-distribution. For 90% confidence, $z_{\alpha /2}=1.645$.

$\frac{\sigma }{\sqrt{n}}=\frac{3.5}{\sqrt{40}}\approx 0.5534$.

$E=1.645\times 0.5534\approx 0.910$.

$120\pm 0.910⟹(119.09,120.91)$.

We are 90% confident the true mean battery lifetime lies between 119.09 and 120.91 hours.

Uses $z=1.96$ (the 95% critical value) instead of 1.645 for 90% confidence.

Uses $z=2.576$ (the 99% critical value) instead of the correct 90% value.

Incorrectly divides $\sigma$ by $n$ instead of $\sqrt{n}$, giving too small a margin of error.

Only option A uses the correct critical value and standard error calculation.

Degrees of freedom for a $t$-interval are $\nu =n-1$.

Here $n=15$, so $\nu =15-1$.

$\nu =14$.

The correct degrees of freedom is 14.

Using $\nu =n=15$ ignores the requirement to subtract 1 for estimating $\mu$.

Adds 1 instead of subtracting 1, a sign error.

Subtracts 2 instead of 1, an over-correction.

Only $\nu =n-1=14$ matches the definition.