Question

A production line produces rulers that are supposed to be 12 inches long. A sample of 49 of the rulers had a mean of 12.1 and a standard deviation of .5 inches. The quality control specialist responsible for the production line decides to do a hypothesis test at the 90 percent significance level to determine whether the production line is really producing rulers that are 12 inches long or not.a. What is the null hypothesis? b. What is the alternative hypothesis? c. What is the value of the test statistic? d. What is the rejection region (with its numerical value)? e. What conclusion do you draw? f. What does this mean in terms of the problem situation?

Answer 1

Answer:

a) Null hypothesis:$\mu =12$  

b) Alternative hypothesis:$\mu \neq 12$  

c) $t=\frac{12.1-12}{\frac{0.5}{\sqrt{49}}}=1.4$  

d) $t_{cric}=\pm 1.68$

And the rejection region would be $(-\infty, -1.68) \cup (1.68, \infty)$

e) Since the calculated value is not on the rejection zone we can't reject the null hypothesis

f)For this case we can conclude at 10% of significance that the true mean is not significantly different from 12 and then the specification is not violated.

Step-by-step explanation:

Data given and notation  

$\bar X=12.1$ represent the sample mean

$s=0.5$ represent the sample deviation

$n=49$ sample size  

$\mu_o =12$ represent the value that we want to test  

$\alpha=0.1$ represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

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

Part a

We need to conduct a hypothesis in order to check if the mean is equal to 12 or not, the system of hypothesis would be:  

Null hypothesis:$\mu =12$  

Part b

Alternative hypothesis:$\mu \neq 12$  

Part c

The statistic is given by:

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

$t=\frac{12.1-12}{\frac{0.5}{\sqrt{49}}}=1.4$  

Part d

For this case since we are conducting a bilateral test we need to find the degrees of freedom first given by:

$df = n-1 = 49-1 =48$

And we need to find a critical value in the distribution with 48 degrees of freedom who accumulates $(1-0.9)/2 = 0.05$ of the area and we got:

$t_{cric}=\pm 1.68$

And the rejection region would be $(-\infty, -1.68) \cup (1.68, \infty)$

Part e

Since the calculated value is not on the rejection zone we can't reject the null hypothesis

Part f

For this case we can conclude at 10% of significance that the true mean is not significantly different from 12 and then the specification is not violated.