Question 1086325
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.