A light bulb manufacturer guarantees that the mean life a of a certain type of l
ID: 3397096 • Letter: A
Question
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)Explanation / Answer
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.
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