SOLUTION: A firm that produces light bulbs claims that their lightbulbs last 1500 hours, on average. You wonder if the average might differ from the 1500 hours that the firm claims. To expl

Question 783977: A firm that produces light bulbs claims that their lightbulbs last 1500 hours, on average. You wonder if the average might differ from the 1500 hours that the firm claims. To explore this possibility you take a random sample of n = 25 light bulbs purchased from this firm and record the lifetime (in hours) of each bulb. You then conduct an appopriate test of hypothesis. Some of the information related to the hypothesis test is presented below.
Test of H_0: = 1500 versus H_1: ≠ 1500
Sample mean 1509.5 Std error of mean 4.854
Assuming the life length of this type of light bulb is normally distributed, if you wish to conduct this test using a .05 level of significance, what are the critical values that you should use?
Answer by ed42ptt(4) (Show Source): You can put this solution on YOUR website!
(X ̅-μ)/(σ/√n)=>Standard Error=σ/√n=(24.27)/√25=4.854;
Thus using (X ̅-μ)/E would be the new formula to use (�1509.5-1500)/(4.854)=>(�9.5)/(4.854)≈
(-1.957 and 1.957) this gives you the t-values not the critical values.
In addition, the p-value for 2 tails would be 0.062. Using the free p-value online calculator. On the TI 83/84 calculator enter 2nd distr number 5 tcdf(1.957, 1000, 24) = 0.031*2=0.062 > 0.05
Critical value pertaining to this test is found in Table F, t-distribution table, at 24 degrees of freedom, 0.05, and 2 tails you find that the values will be between -2.093 < 2.093 thus TI83/84 will show the p-value of 0.05 such that 2nd distr, number 5 tcdf(2.093, 1000, 24)=0.0250*2 = 0.050 which relates to the original hypothesis in the question.

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