SOLUTION: An engineer claims that her new battery will operate continuously for at least 7 minutes longer than the old battery. To test the claim, the company selects a simple random sample

Question 960379: An engineer claims that her new battery will operate continuously for at least 7 minutes longer than the old battery. To test the claim, the company selects a simple random sample of 100 new batteries and 100 old batteries. The old batteries run continuously for 190 minutes with a standard deviation of 25 minutes; the new batteries, 200 minutes with a standard deviation of 35 minutes. Test the engineer's claim that the new batteries run at least 7 minutes longer than the old. Use a 0.05 level of significance.
A) state the null and alternative hypothesis, B) determine the critical value(s) that define the rejection region, C) calculate the value of the test statistic, D) make the decision to reject the null hypothesis or not, and E) state a full conclusion, in context

Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website!
A) Ho = The new batteries run at least 7 minutes longer than the old batteries.
Ha = The new batteries do not run at least 7 minutes longer than the old batteries.
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B) We expect the new batteries to have a mean > or = 197 ( at least 7 more than the mean for old batteries). Now we can calculate the z-value for the new batteries,
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C) z-value = (197 - 200) / (35/10) = -3 / 3.5 = −0.857142857 approx -0.86,
the associate p-value is 0.1949
note that Probability(X>197) = 1 - Probability(X<197) = 1 - 0.1949 = 0.8051
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D) we are asked to use the .05 significance level, therefore
the p-value of 0.1949 > .05 so we accept the null hypothesis
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E) Because 200 is consistent with the null hypothesis (� is greater than or equal to 197), hence inconsistent with the alternative hypothesis (� < 197); you do not reject the null hypothesis.

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