Question 938368: The ABC battery company claims that their batteries last at least 100 hours, on average. Your experience with their batteries has been somewhat different, so you decide to conduct a test to see if the company's claim is true. You believe that the mean life is actually less than the 100 hours the company claims. You decide to collect data on the average battery life (in hours) of a random sample of n = 20 batteries. Some of the information related to the hypothesis test is presented below.
Test of H0: 100 versus H1: 100
Sample mean 98.5
Std error of mean 0.777
Assuming the life length of batteries is normally distributed, what is the p-value associated with this test? Place your answer, rounded to 3 decimal places in the blank. For example, 0.234 would be a legitimate entry
Found 2 solutions by ewatrrr, ed42ptt:
Answer by ewatrrr(24785)   (Show Source):
You can put this solution on YOUR website! Ho: u ≥ 100 (claim)
Ha: u < 100
..
20 Sample: mean = 98.5, s = .777
...
test stat  : t = -1.5/(.777/sqrt(20))
p = p(t < -8.6335), which is basically 0
....
Claim rejected
Answer by ed42ptt(4)  (Show Source):
You can put this solution on YOUR website! (X ̅-μ)/(σ/√n)=>Standard Error=σ/√n=(3.475)/√20=0.777;
Thus using (X ̅-μ)/E would be the new formula to use (±98.5-100)/(0.777)=>(±1.5)/(0.777)≈
(-1.931 and 1.931) are the t-values associated with finding the p-value. Once you have the t-value, you can find the p-value by using the ti-83/84 2nd VARS(dist, number 5 tcdf(1.931,1000,19) = 0.0343*2 = 0.0686 > 0.01 > 0.05 < 0.10
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