Respuesta :
Answer:
[tex]z=\frac{112-124}{\frac{22}{\sqrt{37}}}=-3.318[/tex]
The p value would be given by this probability:
[tex]p_v =2*P(z<-3.318)=0.0009[/tex]
Since the p value is a very small value at any significance level used we can reject the null hypothesis and we can conclude that the true mean for this case is different from 124 ft
Step-by-step explanation:
Data given and notation
[tex]\bar X=112[/tex] represent the sample mean
[tex]\sigma =22[/tex] represent the population standard deviation
[tex]n=37[/tex] sample size
[tex]\mu_o =124[/tex] represent the value that we want to test
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value
Hypothesis to test
We want to check the following system of hypothesis:
Null hypothesis: [tex]\mu = 124[/tex]
Alternative hypothesis :[tex]\mu \neq 124[/tex]
The statistic is given by:
[tex]z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}[/tex] (1)
Replacing the info given we got:
[tex]z=\frac{112-124}{\frac{22}{\sqrt{37}}}=-3.318[/tex]
The p value would be given by this probability:
[tex]p_v =2*P(z<-3.318)=0.0009[/tex]
Since the p value is a very small value at any significance level used we can reject the null hypothesis and we can conclude that the true mean for this case is different from 124 ft