Respuesta :
Answer:
The null hypothesis is [tex]H_0: \mu = 14[/tex]
The alternate hypothesis is [tex]H_a: \mu < 14[/tex]
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:
[tex]H_0: \mu = 14[/tex]
At the alternate hypothesis, we test if the mean is less than 14 hours, that is:
[tex]H_a: \mu < 14[/tex]
The test statistic is:
[tex]t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}[/tex]
In which X is the sample mean, [tex]\mu[/tex] 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 [tex]\mu = 14[/tex]
X = 13.6 hours, s = 1.3 hours. Sample of 40:
In addition to the values of X and s given, we have that [tex]n = 40[/tex]
Test statistic:
[tex]t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}[/tex]
[tex]t = \frac{13.6 - 14}{\frac{1.3}{\sqrt{40}}}[/tex]
[tex]t = -1.95[/tex]
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.
