Question

A random sample of ?n measurements was selected from a population with standard deviation 13.913.9 and unknown mean ?μ. Calculate a 9999 % confidence interval for ?μ for each of the following situations:

(a) ?=35, ?⎯⎯⎯=104.9n=35, x¯=104.9

(b) ?=60, ?⎯⎯⎯=104.9n=60, x¯=104.9

(c) ?=85, ?⎯⎯⎯=104.9n=85, x¯=104.9

(d) In general, we can say that for the same confidence level, increasing the sample size  the margin of error (width) of the confidence interval. (Enter: ''DECREASES'', ''DOES NOT CHANGE'' or ''INCREASES'', without the quotes.)

Answer 1

Ans (a)

99% confidence interval for when =104.9,n=35 , =13.9

At 99% CI the value of Z is 2.58

ME= margin of error

ME=Z*/

ME=2.58*13.9/

ME=6.0617

99%confidence interval is

(- ME, +Me)

(98.8382 ,110.9617)

Similarly, we can find b and c

b). n= 60

ME=4.0914

99% confidence interval is

(100.2702 ,109.5297)

C) n=85

ME=3.8897

99% confidence interval is

(101.0102 , 108.7897)

#conclusion : we can say that for the same confidence level ,increasing the sample size ,the margin of error ( width) of CI decreases.