Respuesta :
Answer:
[tex]z=\frac{514-600}{\frac{112}{\sqrt{67}}}=-6.285[/tex]
The p value for this case is given by:
[tex]p_v =P(z<-6.285)=1.64x10^{-10}[/tex]
Since the p value is very low than the significance level given we have enough evidence to conclude that the true mean for the scores on a standardized memory is significantly lower than 600 and then we can conclude that memory performance is reduced by old age
Step-by-step explanation:
Information given
[tex]\bar X=514[/tex] represent the sample mean for the scores on a standardized memory
[tex]\sigma=112[/tex] represent the population standard deviation
[tex]n=67[/tex] sample size
[tex]\mu_o =600[/tex] represent the value that we want to verify
[tex]\alpha=0.1[/tex] represent the significance level
z would represent the statistic
[tex]p_v[/tex] represent the p value
System of hypothesis
We want to check if the true mean for this case is less than 600, the system of hypothesis would be:
Null hypothesis:[tex]\mu \geq 600[/tex]
Alternative hypothesis:[tex]\mu < 600[/tex]
Since we know the population deviation the statistic for this case 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{514-600}{\frac{112}{\sqrt{67}}}=-6.285[/tex]
The p value for this case is given by:
[tex]p_v =P(z<-6.285)=1.64x10^{-10}[/tex]
Since the p value is very low than the significance level given we have enough evidence to conclude that the true mean for the scores on a standardized memory is significantly lower than 600 and then we can conclude that memory performance is reduced by old age