Respuesta :
Answer:
[tex]z=\frac{6.6-6.5}{\frac{0.7}{\sqrt{270}}}=2.347[/tex]
The p value for this case would be given by"
[tex]p_v =2*P(z>2.347)=0.0189[/tex]
For this case since the p value is higher than the significance level we don't have enough evidence to conclude that the true mean is significantly different from 6.5 lbs/square inch at 10% of significance. So then there is not enough evidence to conclude that the valve does not perform to the specifications
Step-by-step explanation:
Information given
[tex]\bar X=6.6[/tex] represent the sample mean
[tex]s=\sqrt{0.49}= 0.7[/tex] represent the population deviation
[tex]n=270[/tex] sample size
[tex]\mu_o =6.5[/tex] represent the value that we want to test
[tex]\alpha=0.1[/tex] represent the significance level
z would represent the statistic
[tex]p_v[/tex] represent the p value for the test
Hypothesis to verify
We want to verify if the true mean for this case is equal to 6.5 lbs/square inch or not , the system of hypothesis would be:
Null hypothesis:[tex]\mu= 6.5[/tex]
Alternative hypothesis:[tex]\mu \neq 6.5[/tex]
The statistic for this case is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
And replacing we got:
[tex]z=\frac{6.6-6.5}{\frac{0.7}{\sqrt{270}}}=2.347[/tex]
The p value for this case would be given by"
[tex]p_v =2*P(z>2.347)=0.0189[/tex]
For this case since the p value is higher than the significance level we don't have enough evidence to conclude that the true mean is significantly different from 6.5 lbs/square inch at 10% of significance. So then there is not enough evidence to conclude that the valve does not perform to the specifications