# How do you find a 95% confidence interval for μ when n = 32, x-bar = 137 ft., and σ = 7.0 ft?

Feb 19, 2017

95% C.I. is $137 \text{ ft"+-2.425 " ft}$.

#### Explanation:

Since we want to be 95% confident that our interval contains the true value of $\mu$, we set $\alpha = 0.05$. Then the 100(1-alpha)% confidence interval for $\mu$ is found by the formula:

bar x +- t_(alpha//2, n–1)sigma/sqrtn

=137 " ft"+-t_(0.025,31)(7.0" ft")/(sqrt 32)

The $t$-value comes from Student's $t$-distribution that has $n - 1$ degrees of freedom; the area under the $t$-curve to the right of that $t$-value is 0.025. The value for t_(alpha//2, n–1) is usually found by lookup in a $t$-table, but for values of $n > 30$, we know t_(alpha//2, n–1) ~~ z_(alpha//2), so we can look up this $z$-value instead. (As $n$ grows larger, $t$ approaches the standard normal distribution $Z$.)

So our 95% confidence interval for $\mu$ becomes

137 " ft"+-z_(0.025)(7.0" ft")/(sqrt 32)

~~137 " ft"+-(1.96)(1.237" ft")

$\approx 137 \text{ ft"+-2.425 " ft}$.