Respuesta :
Answer:
a) Null hypothesis:[tex]\mu \leq 10[/tex]
Alternative hypothesis:[tex]\mu > 10[/tex]
We don't know the population deviation so then the statistic is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
Replacing the data given we got:
[tex]t=\frac{10.94-10}{\frac{5.10}{\sqrt{49}}}=1.29[/tex]
b) [tex]df=n-1=49-1=48[/tex]
[tex]p_v =P(t_{48}>1.29)=0.102[/tex]
Since the p value is higher than the significance level we have enough evidence to conclude that the true mean for the unemployed workers is significantly higher for the GHQ score
Step-by-step explanation:
Information given
[tex]\bar X=10.94[/tex] represent the sample mean for the GHQ score for unemployed workers
[tex]s=5.10[/tex] represent the sample standard deviation
[tex]n=49[/tex] sample size selected
[tex]\mu_o =10[/tex] represent the limit value for the GHQ score
[tex]\alpha=0.01[/tex] represent the significance level
t would represent the statistic
[tex]p_v[/tex] represent the p value for the test
Part a : Hypothesis to check
We want to determine if unemployed workers tend to be mentally unstable (mean more than 10), the system of hypothesis would be:
Null hypothesis:[tex]\mu \leq 10[/tex]
Alternative hypothesis:[tex]\mu > 10[/tex]
We don't know the population deviation so then the statistic is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
Replacing the data given we got:
[tex]t=\frac{10.94-10}{\frac{5.10}{\sqrt{49}}}=1.29[/tex]
Part b: P value
The degrees of freedom for this test are:
[tex]df=n-1=49-1=48[/tex]
Since the test is right tailed the p value would be:
[tex]p_v =P(t_{48}>1.29)=0.102[/tex]
Since the p value is higher than the significance level we have enough evidence to conclude that the true mean for the unemployed workers is significantly higher for the GHQ score