Question

8.) (15 pts) Construct a 90% lower tailed confidence interval around the true value of fi, when the sample data used are: 45, 37, 65, 64, 46, 60, 45, 34, 48, 51, 55 and 49. 9.) (15 pts) How large of a sample size, n, do you need to be 99% confident that your margin of error is no more than 2 points, when the population standard deviation is (sigma = 15)?

Answer 1

8) Confidence Interval(CI) =

X tdf=11,0=0.1 * S/ vn

$\overline{X}$= Average of all given values

is t statistic value @ degrees of freedom= 11 and significance level=0.1 This will be obtained from t distribution table

tdf=11,a=0.1

S = Sample Standard deviation

n = Sample size

	Data
	45
	37
	65
	64
	46
	60
	45
	34
	48
	51
	55
	49
Sum	599
Mean	49.92
Standard Deviation	9.74
Sample size(n)	12
Confidence Level(CL)	90%
Significance level(alpha)	0.1
Degrees of fredom(n-1)	11
t statistic @ 90% CI,one tail	1.36
Standard error	2.81
Marginal Error	3.83
	Lower CI	Upper CI
Confidence Interval(CI)	46.08	53.75

9) Z critical value for 99% CI or 0.01 significance level is 2.58 (considering Two tailed).

Margin of error =

Za=0.005 * S/n

Therefore

n = (Za=0.005 * S/ME)

Given

S = 15

ME < 2

By substituting the values in the above formula we will get n value which is 305(Sample size)