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The quality-control manager at a light bulb factory needs to determine whether the mean life of a large shipment

of light bulbs is equal to 375 hours.  The population standard deviation is 120 hours.  A random sample of 64 light bulbs indicates a sample mean life of 350 hours.
A.  At the 0.05 level of significance, is there evidence that the mean life is different from 375 hours?
B.  Compute the p-value and interpret its meaning.
C.  Construct a 95% confidence interval estimate of the population mean life of the light bulbs.
D.  Compare the results of (a) and (c).  What conclusions do you reach?
E.  Compare the results of (a) if the standard deviation were 100 hours instead of 120 hours.

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Karla R. | Let Me Help You With Your Math!(Elem Math thru Pre-Calc and SAT Math)Let Me Help You With Your Math!(Elem Mat...
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H0 :  μ= 375
H: μ ≠ 375 (claim)
α = 0.05 (significance level)
α/2= 0.025(two-tailed)
A. Since the key word "different" is the question, then this is a two-tailed test. Therefore, divide 0.05 by 2 = .025(per tail). We find the critical values by going to the z-score tables for normal distribution. Notice that the area= .025 corresponds to z-scores(critical values)= ±1.96. To calculate test-statistic manually: (350-375)/(120/√64) =-25/15= -1.66 . Since the test-statistic does not land within the critical values region, then we can conclude  at a 0.05 significance level that there is  not sufficient evidence to support the claim that the mean life is different from 375 hours.
B. Using Graphing calculator TI-83: go to STATS, TESTS, Z-Test, Stats, μ0:375, σ:120, x:350, n:64, μ: ≠μ0,  CALCULATE. p-value=.096 which is greater than α=0.05. Therefore we support the null hypothesis(H0)and reject the claim of the alternative hypothesis(H1).
C .Using graphing calculator TI-83: STATS, TESTS,7↓ZInterval...,Stats, σ:120, x:350, n:64, C-Level:.95,          Calculate= (320.6, 379.4)
D. The TRUE mean lies within the interval. So, 375 does lie between 320.6 and 379.4, and, therefore, still could be 375.
E. Entering in σ=100 in calculations above (instead of σ=0), we get p-value=0.04 which is less than α=0.05. Therefore, we reject the null, H0, and support the claim(H1) that the mean life is different from 375. Also, 375 was not within the 95%CI = (325.5,374.5).