Respuesta :
Answer:
[tex]t=\frac{435-430}{\frac{29}{\sqrt{23}}}=0.827[/tex]
The degrees of freedom are given by:
[tex]df=n-1=23-1=22[/tex]
And the p value taking in count that we have a bilateral test we got:
[tex]p_v =2*P(t_{(22)}>0.827)=0.417[/tex]
Since the p value is higher than the significance level of 0.05 we have enough evidence to conclude that the true mean is not significantly different from 430 the required value
Step-by-step explanation:
Information given
[tex]\bar X=435[/tex] represent the mean for the weight
[tex]s=\sqrt{841}=29[/tex] represent the sample standard deviation
[tex]n=23[/tex] sample size
[tex]\mu_o =430[/tex] represent the value that we want to verify
[tex]\alpha=0.05[/tex] represent the significance level
t would represent the statistic
[tex]p_v[/tex] represent the p value
System of hypothesis
We are trying to proof if the filling machine works correctly at the 430 gram setting, so then the system of hypothesis for this case are:
Null hypothesis:[tex]\mu = 430[/tex]
Alternative hypothesis:[tex]\mu \neq 430[/tex]
In order to cehck the hypothesis the statistic for a one sample mean test is given by
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
Replacing the info given we have this:
[tex]t=\frac{435-430}{\frac{29}{\sqrt{23}}}=0.827[/tex]
The degrees of freedom are given by:
[tex]df=n-1=23-1=22[/tex]
And the p value taking in count that we have a bilateral test we got:
[tex]p_v =2*P(t_{(22)}>0.827)=0.417[/tex]
Since the p value is higher than the significance level of 0.05 we have enough evidence to conclude that the true mean is not significantly different from 430 the required value
