Interesting statistics question...any ideas?

Question

An economist is interested in studying the incomes of consumers in a particular region. The population standard deviation is known to be $1,000. A random sample of 50 individuals resulted in an average income of $15,000. What is the width of the 90% confidence interval?

Answer 1

mean = 15000
s = 1000
n = deg freedom = 50
mu = CI
t= 1.645 (from a Student's t value table)

mu = mean +/- ts/sqrt n
mu = 15000 +/- 3500

EDIT - My bad. I recalculated and have the new answer, +/- 230.

Answer 2

The confidence interval is 15,000 ± z(sub α/2)(σ)/√n
where z(sub α/2) = 1.645 for 90% confidence
We can use the z normal variate (instead of t) because the standard deviation is known.
Evaluating, we just have to find the ± term because the width of the interval is required. One-half the width is:
z(sub α/2)(s)/√n = 1.645(1000)/7.07 = 232.7
So the width is 2x 232.7 = 465.3

Answer 3

25

Answer 4

without having the formula handy, its about 1.9 standard deviations or so , so its about 13,100 to 16, 900 give or take , off the top of my head ( 2 standard deviatio9ns = 95.4 percent or close to that).

there is no way the first answer 1.45 is right because , 68 percent confidence would be 1 standard deviation,( 1000) and the second is about 27 percent more ( 68 to 95 confidence) so 90 percent confidence would be closer to 2 standard deviations, 90 to 95, than to one 68 to 90, hence it wouldnt be 1.45 has to be greater..........

Answer 5

the confidence interval is given by
mean+Z*std dev
where Z is the value of variate at that level of significance...

since Z=0.1736 at ..1 level of significance (as the normal distribution is symmetric) so the width will be equal to 2*Z*1000=2*0.1736*1000 =347.2