Respuesta :
Answer:
[tex]z=\frac{19.87-20}{\frac{0.4}{\sqrt{25}}}=-1.625[/tex]
The p value for this case would be given by:
[tex]p_v =P(z<-1.625)=0.052[/tex]
Since the p value is higher than the significance level provided we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly less than 20 ounces.
Step-by-step explanation:
Information provided
[tex]\bar X=19.87[/tex] represent the sample mean
[tex]\sigma=0.4[/tex] represent the population deviation
[tex]n=25[/tex] sample size
[tex]\mu_o =68[/tex] represent the value that we want to test
[tex]\alpha=0.01[/tex] represent the significance level
z would represent the statistic
[tex]p_v[/tex] represent the p value
Hypothesis to test
We want to test if the true mean is at least 20 ounces, the system of hypothesis would be:
Null hypothesis:[tex]\mu \geq 20[/tex]
Alternative hypothesis:[tex]\mu < 20[/tex]
The statistic 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{19.87-20}{\frac{0.4}{\sqrt{25}}}=-1.625[/tex]
The p value for this case would be given by:
[tex]p_v =P(z<-1.625)=0.052[/tex]
Since the p value is higher than the significance level provided we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly less than 20 ounces.
