Question

Bob is the owner of a home improvement store. He has hired you to check his machine’s calibration prior to starting production on a large order. To check this, you set the machine to create 1.5 inch bolts and manufacture a random sample of 200 bolts. That sample of bolts has an average length of 1.521 inches with a standard deviation of 0.204 inches. Does this sample provide convincing evidence that the machine is working properly or should it be shut down for repairs?

Answer 1

Answer:

$t=\frac{1.521-1.5}{\frac{0.204}{\sqrt{200}}}=1.456$  

$p_v =2*P(z>1.456)=0.145$  

If we compare the p value and the significance level assumed $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL reject the null hypothesis, so we can conclude that the true mean is not different from the specification at 1% of signficance.  

Step-by-step explanation:

Data given and notation  

$\bar X=1.521$ represent the sample mean

$s=0.204$ represent the sample standard deviation  

$n=200$ sample size  

$\mu_o =1.5$ represent the value that we want to test  

$\alpha$ represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

$p_v$ represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is different from 1.5 inch, the system of hypothesis would be:  

Null hypothesis:$\mu =1.5$  

Alternative hypothesis:$\mu \neq 1.5$  

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic  

We can replace in formula (1) the info given like this:  

$t=\frac{1.521-1.5}{\frac{0.204}{\sqrt{200}}}=1.456$  

P-value  

The degrees of freedom are given by:

$df = n-1= 200-1=199$

Since is a two-sided test the p value would be:  

$p_v =2*P(z>1.456)=0.145$  

Conclusion  

If we compare the p value and the significance level assumed $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL reject the null hypothesis, so we can conclude that the true mean is not different from the specification at 1% of signficance.