Answer :
Answer:
[tex]z=\frac{931-1131}{\frac{333}{\sqrt{30}}}=-3.290[/tex]
Now we can calculate the p value using the alternative hypothesis:
[tex]p_v =P(z<-3.290)=0.0005[/tex]
Since the p value is lower than the significance level of 0.05 we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly less than 1131
Step-by-step explanation:
Information given
[tex]\bar X=931[/tex] represent the sample mean
[tex]\sigma=333[/tex] represent the population standard deviation
[tex]n=30[/tex] sample size
[tex]\mu_o =1131[/tex] represent the value to check
[tex]\alpha=0.05[/tex] represent the significance level
z would represent the statistic
[tex]p_v[/tex] represent the p value
System of hypothesis
We want to verify if the true mean is less than 1131, the system of hypothesis would be:
Null hypothesis:[tex]\mu \geq 1131[/tex]
Alternative hypothesis:[tex]\mu <1131[/tex]
The statistic is given by:
[tex]z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}[/tex] (1)
Reaplacing we got:
[tex]z=\frac{931-1131}{\frac{333}{\sqrt{30}}}=-3.290[/tex]
Now we can calculate the p value using the alternative hypothesis:
[tex]p_v =P(z<-3.290)=0.0005[/tex]
Since the p value is lower than the significance level of 0.05 we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly less than 1131