Respuesta :
Answer:
The pvalue of 0.0113 < 0.05 means that there is sufficient evidence to conclude that the mean time to find another position is less than 28 weeks at the 5% level of significance
Step-by-step explanation:
The null hypothesis is:
[tex]H_{0} = 28[/tex]
The alternate hypotesis is:
[tex]H_{1} < 28[/tex]
The test statistic is:
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
In which X is the sample mean, [tex]\mu[/tex] is the value tested at the null hypothesis, [tex]\sigma[/tex] is the standard deviation and n is the size of the sample.
A recent survey of 50 executives who were laid off during a recent recession revealed it took a mean of 26 weeks for them to find another position.
This means that [tex]n = 50, X = 26[/tex]
Assume the population standard deviation is 6.2 weeks.
This means that [tex]\sigma = 6.2[/tex]
Does the data provide sufficient evidence to conclude that the mean time to find another position is less than 28 weeks at the 5% level of significance
We have to find the pvalue of Z, looking at the z-table, when [tex]\mu = 28[/tex]. It if is lower than 0.05, it provides evidence.
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]z = \frac{26 - 28}{\frac{6.2}{\sqrt{50}}}[/tex]
[tex]z = -2.28[/tex]
[tex]z = -2.28[/tex] has a pvalue of 0.0113 < 0.05.
The pvalue of 0.0113 < 0.05 means that there is sufficient evidence to conclude that the mean time to find another position is less than 28 weeks at the 5% level of significance
