Question 318653: The Simmons furniture Company selected a random sample of 36 sofas that were made at its Memphis factory. Each sofa was subjected to a test process that simulated people sitting on the cushions. The test requires that a heavy object be repeatedly dropped on the cushion until the fabric wears out. 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.
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.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! 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.
---
x-bar = 12600
ME = (invT(0.975 with df=35)*1900/sqrt(36) = 2,0301*316.67 = 642.87
---
95% CI: 12600-642.87 < u < 12600+642.87
95% CI: 13242.87 < u < 13242.87
=======================================================
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.
---
# of drops = 20*365*2 = 14600
====
No, Simmons is not justified. 14600 is not in the 95% confidence interval.
========================
Cheers,
Stan H.
========================
|
|
|
