Question

2. A manufacturer of transistors claims that its transistors will last an average of 1000 hours. To maintain this claim, 25 resistors are tested each month. What conclusions can you draw from a sample that has a mean of 1010 and a standard deviation of 60? You may assume that the distribution of the lifetime of a transistor is normal [NOTE: To receive FULL CREDIT, you must conduct a COMPLETE hypothesis test and make your conclusions using BOTH a fixed O.01 significance level and a p-value analusis.] a. State your null and alternative hypotheses: b. What is the appropriate test statistic for this hypothesis test? c. What are the critical values for this test and what is the decision criteria - i.e. when will you reject the null hypothesis? (fixed and p- value) d. What is the value of the test statistic? e. What is the p-value for this test? Based upon your analysis of the above, clearly state your DECISION for this hypothesis test and your rationale. f.

Answer 1

Part a)

H0 :-

H1 :- $\mu \neq 1000$

Part b)

We can use t test, since we have sample standard deviation

Part c)

Test Criteria :-

$Reject null hypothesis if | t |\; > \;t_{\alpha /2, n-1}$

$t_{\alpha /2, n-1} = t_{\0.01 /2, 25-1} = 2.797$

$| t |\; > \;t_{\alpha /2, n-1} = 0.8333 < 2.797$

Part d)

Test Statistic :-

$t = ( \bar{X} - \mu ) / (S /\sqrt{n})$

$t = ( 1010 - 1000 ) / ( 60 /\sqrt{ 25 })$

t = 0.8333

Part e)

P value

Looking for t = 0.8333 in t table across 24 degree of freedom

t = 0.8333 lies between the value 0.685 and 0.857 with respective P value are 0.50 and 0.40.

Using excel to calculate exact P value = 0.4129

Part f)

Decision based on P value

Reject null hypothesis if P value < $\alpha=0.01$ level of significance

0.4129 > 0.01, we fail to reject null hypothesis

Decision based on critical region

$t_{\alpha /2, n-1} = t_{\0.01 /2, 25-1} = 2.797$

$| t |\; > \;t_{\alpha /2, n-1} = 0.8333 < 2.797$

Result :- Fail to reject null hypothesis