# Test null hypothesis that the mean value is less than 60

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.

• 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. Nov 28 '20 at 20:32

## 2 Answers

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)


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.