Answer:
Does this sample provide convincing evidence that the machine is working properly?
Yes.
Step-by-step explanation:
Normal distribution test:
[tex]$z=\frac{x- \mu }{ \frac{\sigma}{\sqrt{n}} }=\frac{ (x-\mu)\sqrt{n}}{\sigma} $[/tex]
Where,
[tex]x: \text{ sample mean}[/tex]
[tex]\sigma: \text{ standard deviation}[/tex]
[tex]n: \text{ sample size determination}[/tex]
[tex]\mu: \text{ hypothesized size of the screw}[/tex]
[tex]$z=\frac{(1.476-1.5)\sqrt{200} }{0.203 } $[/tex]
[tex]$z=\frac{(-0.024)10\sqrt{2} }{0.203 } $[/tex]
[tex]z \approx -1.672[/tex]
Once the significance level was not given, It is usually taken an assumption of a 5% significance level.
Taking the significance level of 5%, which means a confidence level of 95%, we have a z-value of [tex]\pm 1.96[/tex]
Therefore, we fail to reject the null. It means that the hypothesis test is not statistically significant: the average length is not different from 1.5!
