Answer:
[tex]z=\frac{256-266}{\frac{16}{\sqrt{20}}}=-2.795[/tex] Â Â
The p value for this case can be calculated with this probability:
[tex]p_v =P(z<-2.795)=0.0026[/tex] Â
Step-by-step explanation:
Information provided
[tex]\bar X=256[/tex] represent the sample mean
[tex]\sigma=16[/tex] represent the population standard deviation
[tex]n=20[/tex] sample size Â
[tex]\mu_o =266[/tex] represent the value that we want to test
z would represent the statistic
[tex]p_v[/tex] represent the p value
Hypothesis to verify
We need to conduct a hypothesis in order to see if inadequate food intake should have shorter gestation times, the system of hypothesis would be: Â
Null hypothesis:[tex]\mu \geq 266[/tex] Â
Alternative hypothesis:[tex]\mu < 266[/tex] Â
The statistic is given by:
[tex]z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}[/tex] Â (1) Â
Replacing the info given we got:
[tex]z=\frac{256-266}{\frac{16}{\sqrt{20}}}=-2.795[/tex] Â Â
The p value for this case can be calculated with this probability:
[tex]p_v =P(z<-2.795)=0.0026[/tex] Â