Respuesta :
Answer:
[tex]z=\frac{16.2-14.5}{\frac{7}{\sqrt{100}}}=2.429[/tex]
The p value for this case is given by:
[tex]p_v =2*P(z>2.429)=0.015[/tex]
The p value for this case is lower than the significance level so then we can conclude that the true mean is significantly different from 14.5 months
Step-by-step explanation:
Information given
[tex]\bar X=16.2[/tex] represent the sample mean
[tex]\sigma=7[/tex] represent the population standard deviation
[tex]n=100[/tex] sample size
[tex]\mu_o =14.5[/tex] represent the value that we want to check
[tex]\alpha=0.05[/tex] represent the significance level
z would represent the statistic
[tex]p_v[/tex] represent the p value for the test
System of hypothesis
We want to check if the true mean is different from 14.5, the system of hypothesis would be:
Null hypothesis:[tex]\mu = 14.5[/tex]
Alternative hypothesis:[tex]\mu \neq 14.5[/tex]
The statistic is given:
[tex]z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}[/tex] (1)
The statistic is given by:
[tex]z=\frac{16.2-14.5}{\frac{7}{\sqrt{100}}}=2.429[/tex]
The p value for this case is given by:
[tex]p_v =2*P(z>2.429)=0.015[/tex]
The p value for this case is lower than the significance level so then we can conclude that the true mean is significantly different from 14.5 months
