Question 1086325: Samsung wants to know how long cell phone owners keep their phones before upgrading. A simple random sample of 23 cell phone owners results in a mean of 2.64 years and a standard deviation of 0.71 years. Assume the sample is drawn from a normally distributed population.
Find the 95% confidence interval of the population mean.
If you worked for Samsung and decided you wanted to be 99% confident that the sample mean is within 0.25 years of the population mean, how large of a sample would you need to take? Assume that σ=0.71 for this calculation.
Explain why the population parameter may NOT follow a normal distribution. Would you expect the data to show a positive or negative skew? Explain. If the data were not normally distributed, how would this affect the calculations for the confidence interval?
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! The 95% CI is mean +/-t *s/sqrt(n)
t df=23, 0.975=2.07
2.64 +/- 2.07*0.71/4.80; the interval width is 0.31
(2.33, 2.95) units years
within 0.25 years
0.25=z*0.71/ sqrt (n);use z until there is an estimate of sample size. z(0.995)=2.576
0.25=1.83/sqrt (n)
cross multiply and square both sides
n=(1.83/0.25)^2=53.6
Need to use t,start with df=54
2.67*0.71/sqrt(54)=0.2579
try df=60, and interval is+/-0.25
NOTE: I'm using a t.
If sigma is known,then the sample size is 54 as shown above.
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I wold expect a positive skew, since people will continue to have their phones at 3,4, 5 years. If the data are not normally distributed, the confidence interval will be wider for a given sample size, but more importantly, another test would be needed, perhaps a non-parametric or distribution-free test.
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