Answer:
Null hypothesis:[tex]\mu \geq 42.3[/tex]
Alternative hypothesis:[tex]\mu < 42.3[/tex]
[tex]t=\frac{40.6-42.3}{\frac{2.7}{\sqrt{24}}}=-3.085[/tex]
The degrees of freedom are given by:
[tex]df=n-1=24-1=23[/tex]
Then the p value can be calculated with this probability:
[tex]p_v =P(t_{(23)}<-3.085)=0.0026[/tex]
If we compare the p value and the significance lvel of 0.1 we see that the p value is lower than the signficance level we can reject the null hypothesis and we can conclude that the the assembly time using the new method is faster
Step-by-step explanation:
Information given
[tex]\bar X=40.6[/tex] represent the sample mean for the new method
[tex]s=2.7[/tex] represent the sample standard deviation
[tex]n=24[/tex] sample size
[tex]\mu_o =42.3[/tex] represent the value that we want to test
[tex]\alpha=0.1[/tex] represent the significance level
t would represent the statistic
[tex]p_v[/tex] represent the p value
System of hypothesis
We want to verify if the assembly time using the new method is faster, the system of hypothesis would be:
Null hypothesis:[tex]\mu \geq 42.3[/tex]
Alternative hypothesis:[tex]\mu < 42.3[/tex]
The statistic for this case is given:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
The statistic would be given by:
[tex]t=\frac{40.6-42.3}{\frac{2.7}{\sqrt{24}}}=-3.085[/tex]
The degrees of freedom are given by:
[tex]df=n-1=24-1=23[/tex]
Then the p value can be calculated with this probability:
[tex]p_v =P(t_{(23)}<-3.085)=0.0026[/tex]
If we compare the p value and the significance lvel of 0.1 we see that the p value is lower than the signficance level we can reject the null hypothesis and we can conclude that the the assembly time using the new method is faster