Respuesta :
Answer:
[tex]z=\frac{(319.5-298.9)-0}{\sqrt{\frac{2.3^2}{59}+\frac{2.0^2}{40}}}}=47.302[/tex]
Thsis value is very high and if we find the p value we would get:
[tex] p_v= P(z>47.302) \approx 0[/tex]
So then we have enough evidence to support the claim that the true mean for males is higher than the true mean for females.
Step-by-step explanation:
Data given and notation
[tex]\bar X_{m}=319.5[/tex] represent the mean for the sample male
[tex]\bar X_{f}=298.9[/tex] represent the mean for the sample female
[tex]\sigma_{m}=2.3[/tex] represent the population standard deviation for the males
[tex]s_{f}=2.0[/tex] represent the population standard deviation for the females
[tex]n_{m}=59[/tex] sample size for the group male
[tex]n_{f}=40[/tex] sample size for the group female
z would represent the statistic (variable of interest)
System of hypothesis
We need to conduct a hypothesis in order to check if that mean scores on the science assessment for male high school students were higher than for the female high school students, the system of hypothesis would be:
Null hypothesis:[tex]\mu_{m}-\mu_{f} \leq 0[/tex]
Alternative hypothesis:[tex]\mu_{m} - \mu_{f}> 0[/tex]
The staistic for this case is:
[tex]z=\frac{(\bar X_{m}-\bar X_{f})-\Delta}{\sqrt{\frac{\sigma^2_{m}}{n_{m}}+\frac{\sigma^2_{f}}{n_{f}}}}[/tex] (1)
With the info given we can replace in formula (1) like this:
[tex]z=\frac{(319.5-298.9)-0}{\sqrt{\frac{2.3^2}{59}+\frac{2.0^2}{40}}}}=47.302[/tex]
Thsis value is very high and if we find the p value we would get:
[tex] p_v= P(z>47.302) \approx 0[/tex]
So then we have enough evidence to support the claim that the true mean for males is higher than the true mean for females.
