Question

A lathe is set to cut bars of steel into lengths of 6 cm. The lathe is considered to be in perfect adjustment if the average length of the bars it cuts is 6 cm. A sample of 93 bars is selected randomly and measured. It is determined that the average length of the bars in the sample is 5.97 cm. The sample standard deviation is 0.4 cm. Determine whether or not the lathe is in perfect adjustment. Use a .05 level of significance and choose the correct test statistic.
A. -0.813
B. -0.831
C. -0.76
D. -0.773
E. -0.723

Answer 1

Answer:

$t=\frac{5.97-6}{\frac{0.4}{\sqrt{93}}}=-0.723$    

E. -0.723

$df=n-1=93-1=92$  

$p_v =2*P(t_{(92)}$  

Since the p value is very high we don't have enough evidence to conclude that the true mean for the lengths is different from 6 cm.

Step-by-step explanation:

Information provided

$\bar X=5.97$ represent the sample mean for the length

$s=0.4$ represent the sample standard deviation

$n=93$ sample size  

$\mu_o =6$ represent the value that we want to test

$\alpha=0.05$ represent the significance level

t would represent the statistic  

$p_v$ represent the p value for the test

System of hypothesis

We need to conduct a hypothesis in order to check if the lathe is in perfect adjustment (6cm), then the system of hypothesis would be:  

Null hypothesis:$\mu = 6$  

Alternative hypothesis:$\mu \neq 6$  

since we don't know the population deviation the statistic is:

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

Replacing in formula (1) we got:

$t=\frac{5.97-6}{\frac{0.4}{\sqrt{93}}}=-0.723$    

E. -0.723

P value

The degrees of freedom are given by:

$df=n-1=93-1=92$  

Since is a two tailed test the p value would be:  

$p_v =2*P(t_{(92)}$  

Since the p value is very high we don't have enough evidence to conclude that the true mean for the lengths is different from 6 cm.