SOLUTION: 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 the

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Question 342066: 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?


Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
A manufacturer claims that the mean lifetime,u , of its light bulbs is 51 months. The standard deviation of these lifetimes is 7 months.
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Sixty bulbs are selected at random, and their mean lifetime is found to be 53 months.
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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?
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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: u = 51
The alternative hypothesis: u is not equal to 51
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:
t = (53-51)/[7/sqrt(60)] = 2.213
Can we conclude that the mean lifetime of the bulbs made by this manufacture differ from 51 months?
p-value = 2*P(t > 2.213 when dr = 59) = 2*tcdf(2.213,100,59) = 2*0.0154
= 0.0308
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Since the p-value is greater than 1%, fail to reject Ho.
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Cheers,
Stan H.