Test null hypothesis that the mean value is less than 60

Question

Test $H_0$ : $\mu >= 60$ using 10% significance level if $s = 16$, $\bar{x} = 66$ and $n = 13$. Do not forget to specify $H_1$.

So, $H_1 : \mu < 60$. With df = 12 t value for 10% significance interval is equal to 1.782. Then standard error: $1.782 \cdot \frac{16}{\sqrt{13}} = 4.438$. And the critical value would be $66 - 1.1782 \cdot 4.438$. So with 90% chance the mean value will lie below that number.I know i've done it incorrectly.

Answer 1

Intuitively, the estimated $\bar x=66$ based on 13 observations is above $60$ when testing $\mu<60$ (not even lower), so there can't be a rejection.

Technically, the steps are:

1. Setup the Hypotheses $H_0, H_1$ and the significance level $\alpha$, which you have correctly done.
2. Calculate the test statistic. This is simply any random variable whose distribution is known if $H_0$ is true. We assume that the observations are i.i.d. and hence that $\bar x \overset{a}{\sim}N(\mu,\frac{\sigma^2}{n})$. Therefore, our test statistic is (approx) t-distributed: $t=\frac{\bar x-\mu_0}{\frac{s}{\sqrt{n}}}\sim t(12)$ where $s$ is the sample standard deviation. Here $t=\frac{66-60}{\frac{16}{\sqrt{13}}}=1.352082$
3. Calculate the critical value. Our test is left-tailed, since we want to get the value that fulfills $\mu<60$ in the most extreme cases, were extreme means only happens with $\alpha$ probability conditioned on $H_0$, i.e. only under the assumption that $\mu\ge 60$ is true. In other words, this is the corresponding quantile. The critical value is: $c_\alpha$ where $P(t\le c_\alpha)=\alpha$, here $c_\alpha=-1.356217$
4. Reject if $t$ is more extreme than $c_\alpha$. Here, $t\le c_\alpha$ is checked, since we have a left-tailed test. The condition is not true in this case. Hence, we don't reject $H_0$

In code (R):

## Left-tailed test
# Params
n = 13
alpha = 0.1
s = 16
mu = 60
mu_est = 66

# Plot
curve(dt(x,n-1),xlim=c(-5,5))
abline(v=qt(alpha,n-1),col='red')
t = (mu_est-mu)/(s/sqrt(n))
points(t,0,col='blue',pch=16)

# Test
t < qt(alpha,n-1)

Answer 2

According to your hypothesis setting, you should do a one tail test, and read the value that corresponds to df 12 and one-tail 0.1, which is $1.356$ in the following table.

Besides, the calculation $66−1.1782⋅4.438$ seems vague.