Respuesta :
Answer:
a)
The null hypothesis is [tex]H_0: \mu \leq 10[/tex]
The alternative hypothesis is [tex]H_1: \mu > 10[/tex]
b-1) The value of the test statistic is t = 1.86.
b-2) The p-value is of 0.0348.
Step-by-step explanation:
Question a:
Test if the battery life is more than twice of 5 hours:
Twice of 5 hours = 5*2 = 10 hours.
At the null hypothesis, we test if the battery life is of 10 hours or less, than is:
[tex]H_0: \mu \leq 10[/tex]
At the alternative hypothesis, we test if the battery life is of more than 10 hours, that is:
[tex]H_1: \mu > 10[/tex]
b-1. Calculate the value of the test statistic.
The test statistic is:
We have the standard deviation for the sample, so the t-distribution is used to solve this question
[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 and n is the size of the sample.
10 is tested at the null hypothesis:
This means that [tex]\mu = 10[/tex]
In order to test the claim, a researcher samples 45 units of the new phone and finds that the sample battery life averages 10.5 hours with a sample standard deviation of 1.8 hours.
This means that [tex]n = 45, X = 10.5, s = 1.8[/tex]
Then
[tex]t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}[/tex]
[tex]t = \frac{10.5 - 10}{\frac{1.8}{\sqrt{45}}}[/tex]
[tex]t = 1.86[/tex]
The value of the test statistic is t = 1.86.
b-2. Find the p-value.
Testing if the mean is more than a value, so a right-tailed test.
Sample of 45, so 45 - 1 = 44 degrees of freedom.
Test statistic t = 1.86.
Using a t-distribution calculator, the p-value is of 0.0348.
