Respuesta :
Answer:
We can conclude that the battery life is greater than the 32 hour claim.
Step-by-step explanation:
The null hypothesis is:
[tex]H_{0} = 32[/tex]
The alternate hypotesis is:
[tex]H_{1} > 32[/tex]
Our test statistic is:
[tex]t = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}[/tex]
In which X is the statistic, [tex]\mu[/tex] is the mean, [tex]\sigma[/tex] is the standard deviation and n is the size of the sample.
In this problem, we have that:
[tex]X = 37.8, \mu = 32, \sigma = 10, n = 18[/tex]
So
[tex]t = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]t = \frac{37.8 - 32}{\frac{10}{\sqrt{18}}}[/tex]
[tex]t = 2.46[/tex]
We need to find the probability of finding a mean time greater than 37.8. If it is 5% of smaller(alpha = 0.05.), we can conclude that the battery life is greater than the 32 hour claim.
Probability of finding a mean time greater than 37.8
1 subtracted by the pvalue of z = t = 2.46.
z = 2.46 has a pvalue of 0.9931
1 - 0.9931 = 0.0069 < 0.05
So we can conclude that the battery life is greater than the 32 hour claim.
