Question 318653
The mean number of drops that were recorded for the 36 sofas was 12,600 and the standard deviation was 1,900.
a. Construct a 95% confidence interval estimate for the mean number of drops until the sofa cushions wear out.
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x-bar = 12600
ME = (invT(0.975 with df=35)*1900/sqrt(36) = 2,0301*316.67 = 642.87
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95% CI: 12600-642.87 < u < 12600+642.87
95% CI: 13242.87 < u < 13242.87
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b. Simmons wishes to advertise that the sofa fabric will last at least 20 years. If a sofa is sat upon an average of twice a day, could Simmons justify their proposed advertisement? Explain your answer.
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# of drops = 20*365*2 = 14600
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No, Simmons is not justified.  14600 is not in the 95% confidence interval.
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Cheers,
Stan H.
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