A manufacturer claims that the mean lifetime, u , of its light bulbs is 51 months. The standard deviation of these lifetimes is 7 months. Sixty bulbs are selected at random, and their mean lifetime is found to be 53 months. Can we conclude, at the 0.1 level of significance, that the mean lifetime of light bulbs made by this manufacturer differs from 51 months? Perform a two-tailed test. Then fill in the table below.
Carry your intermediate computations to at least three decimal places, and round your responses as specified in the table. (If necessary, consult a list of formulas.)
the null hypothesis:
The alternative hypotehsis:
The type of test statistic (choose Z, t, Chi-square, or F)
The value of the test statistic (round to at least three decimal places:
Can we conclude that the mean lifetime of the bulbs made by this manufacture differ from 51 months?

We reject  H₀, and conclude thet the mean lifetime of the bulbs differ from 51 month

Step-by-step explanation:

Manufacturing process under control must produce items that follow a normal distribution.

Manufacturer information:

μ  =  51 months     mean lifetime

σ  =  7 months       standard deviation

Sample Information:

x  =  51 months

n  =  60

Confidence Interval  =  90 %

Then significance level  α  =  10 %  α  =  0.1    α/2  =  0,05

Since it is a manufacturing process the distribution is a normal distribution, and with   n = 60 we should use a Z test on two tails.

Then from z- table z(c) for  α = 0,05  is  z(c)  = 1.64

Hypothesis  Test:

Null Hypothesis                              H₀            x  =  μ

Alternative Hypothesis                  Hₐ            x ≠  μ

To calculate z statistics  z(s)

z(s)  =   (  x   -  μ )  / σ /√n

z(s)  =   (  53  -  51 ) / 7 /√60

z(s)  =  2 * 7.746 / 7

z(s)  = 2.213

Comparing  z(s)  and  z(c)

z(s)  >  z(c)   then  z(s)  is in the rejection region

We reject  H₀, and conclude thet the mean lifetime of the bulbs differ from 51 month

42

step-by-step explanation:

256/6=42