Answer:
The null hypothesis is ![H_0: \mu = 14](https://tex.z-dn.net/?f=H_0%3A%20%5Cmu%20%3D%2014)
The alternate hypothesis is ![H_a: \mu < 14](https://tex.z-dn.net/?f=H_a%3A%20%5Cmu%20%3C%2014)
The test statistic is t = -1.95.
The p-value is of 0.0292. This means that for a level of significance of 0.0292 and higher, there is sufficient evidence to conclude that the highlighters wrote for less than 14 continuous hours.
Step-by-step explanation:
Suppose a consumer product researcher wanted to find out whether a highlighter lasted less than the manufacturer's claim that their highlighters could write continuously for 14 hours.
At the null hypothesis, we test if the mean is 14 hours, that is:
![H_0: \mu = 14](https://tex.z-dn.net/?f=H_0%3A%20%5Cmu%20%3D%2014)
At the alternate hypothesis, we test if the mean is less than 14 hours, that is:
![H_a: \mu < 14](https://tex.z-dn.net/?f=H_a%3A%20%5Cmu%20%3C%2014)
The test statistic is:
![t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Cfrac%7Bs%7D%7B%5Csqrt%7Bn%7D%7D%7D)
In which X is the sample mean,
is the value tested at the null hypothesis, s is the standard deviation of the sample and n is the size of the sample.
14 is tested at the null hypothesis:
This means that ![\mu = 14](https://tex.z-dn.net/?f=%5Cmu%20%3D%2014)
X = 13.6 hours, s = 1.3 hours. Sample of 40:
In addition to the values of X and s given, we have that ![n = 40](https://tex.z-dn.net/?f=n%20%3D%2040)
Test statistic:
![t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Cfrac%7Bs%7D%7B%5Csqrt%7Bn%7D%7D%7D)
![t = \frac{13.6 - 14}{\frac{1.3}{\sqrt{40}}}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B13.6%20-%2014%7D%7B%5Cfrac%7B1.3%7D%7B%5Csqrt%7B40%7D%7D%7D)
![t = -1.95](https://tex.z-dn.net/?f=t%20%3D%20-1.95)
The test statistic is t = -1.95.
P-value:
The p-value of the test is the probability of finding a sample mean lower than 13.6, which is a left tailed test, with t = -1.95 and 40 - 1 = 39 degrees of freedom.
Using a calculator, the p-value is of 0.0292. This means that for a level of significance of 0.0292 and higher, there is sufficient evidence to conclude that the highlighters wrote for less than 14 continuous hours.