4
$\begingroup$

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.

$\endgroup$
1
  • 2
    $\begingroup$ You have a very small sample size. Do you know if you're sampling from a normal population? If you don't have this information then I wouldn't perform this hypothesis test. $\endgroup$
    – Matthew H.
    Nov 28 '20 at 20:32
6
$\begingroup$

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)
$\endgroup$
0
2
$\begingroup$

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.

enter image description here

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.