Find the minimum sample size n needed to estimate mu for the given values of c, sigma, and E. c= 0.98, sigma= 8.7, and E= 1
Find the margin of error for the given values of c, s, and n. c=0.98, s=2.4, n =25
Find the critical value Tc for the confidence level c=0.99 and sample size n=88.
Solution :
a ) Given that,
standard deviation = = 8.7
margin of error = E = 1
At 98% confidence level the z is ,
= 1 - 98% = 1 - 0.98 = 0.02
/ 2 = 0.02 / 2 = 0.01
Z/2 = Z0.01 = 2.326
Sample size = n = ((Z/2 * ) / E)2
= ((2.326 * 6250) / 1)2
= 410
Sample size = 410
b ) Given that,
= 3.2
s = 0.9
n = 25
Degrees of freedom = df = n - 1 = 25 - 1 = 24
At 98% confidence level the z is ,
= 1 - 98% = 1 - 0.98 = 0.02
/ 2 = 0.02 / 2 = 0.01
t /2,df = t0.01,24 =2.492
Margin of error = E = t/2,df * (s /n)
= 2.492* (2.4 / 25)
= 1.20
Margin of error = 1.20
c ) Given that,
n = 88
Degrees of freedom = df = n - 1 = 88 - 1 = 87
At 99% confidence level the t is ,
= 1 - 99% = 1 - 0.99 = 0.01
/ 2 = 0.01 / 2 = 0.005
t /2,df = t0.005,87 =2.633
The critcal value t =2.633
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