A light bulb manufacturer guarantees that the mean life a of

A light bulb manufacturer guarantees that the mean life a of a certain type of light bulb is at least 730 hours. A random sample of 25 light bulb has a mean life of 717 hours. Assume the population is normally distributed and the population standard deviation of 63 hours. At alpha = 0.05, do you have enough evidence to reject the manufacture\'s claim Complete parts (a) through (e) Identify the null hypothesis and alternative hypothesis A H_0 : Mu = 717 H_a : 717 (claim) B. H_0Mu = 730 H_8 : Mu = 730 (claim) C. H_0 : 73 (claim) H_8 : Mu = 730 D. H_0 : Mu = 717 H_8 : Mu = 717 (claim) E. H_0: Mu = 717 (claim) H_8 :Mu = 717 F. H_0 : Mu = 730 (claim) H_8 : Mu = 730 Identify the critical values (s). Use technology z_0 Identify the rejection region (s)

Solution

a)

Formulating the null and alternative hypotheses,              
              
Ho:   u   >=   730  
Ha:    u   <   730   [ANSWER, C]

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b)
              
As we can see, this is a    left   tailed test.      
              
Thus, getting the critical z, as alpha =    0.05   ,      
alpha =    0.05          
zcrit =    -   1.644853627   [ANSWER]

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As this is left tailed, then it is OPTION A [CORRECT FIGURE IS OPTION A]

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Other info:

  
Getting the test statistic, as              
              
X = sample mean =    717          
uo = hypothesized mean =    730          
n = sample size =    25          
s = standard deviation =    63          
              
Thus, z = (X - uo) * sqrt(n) / s =    -1.031746032          
              
Also, the p value is              
              
p =    0.151095554          
              
Comparing z and zcrit (or, p and significance level), we   FAIL TO REJECT THE NULL HYPOTHESIS.