Solution
a )Given that,
= 55.3
= 2.2
n = 25
At 95% confidence level the z is ,
= 1 - 95% = 1 - 0.95 = 0.05
/ 2 = 0.05 / 2 = 0.025
Z/2 = Z0.025 = 1.960
Margin of error = E = Z/2* (/n)
= 1.960 * (2.2/ 25 )
= 0.86
At 95% confidence interval estimate of the population mean is,
- E < < + E
55.3 - 0.86 < < 55.3 + 0.86
54.44 < < 56.16
(54.44 , 56.16)
b )Given that,
= 55.3
= 2.2
n = 100
At 95% confidence level the z is ,
= 1 - 95% = 1 - 0.95 = 0.05
/ 2 = 0.05 / 2 = 0.025
Z/2 = Z0.025 = 1.960
Margin of error = E = Z/2* (/n)
= 1.960 * (2.2/ 100 )
= 0.43
At 95% confidence interval estimate of the population mean is,
- E < < + E
55.3 - 0.43< < 55.3 + 0.43
54.87 < < 55.73
(54.87 , 55.73)
c )Given that,
= 55.3
= 2.2
n = 100
At 99% confidence level the z is ,
= 1 - 99% = 1 - 0.99 = 0.01
/ 2 = 0.01 / 2 = 0.005
Z/2 = Z0.005 = 2.576
Margin of error = E = Z/2* (/n)
= 2.576 * (2.2/ 100 )
= 0.57
At 99% confidence interval estimate of the population mean is,
- E < < + E
55.3 - 0.57 < < 55.3 + 0.57
54.73 < < 56.87
(54.73 , 56.87)
d )Given that,
= 55.3
= 2.2
n = 100
At 82% confidence level the z is ,
= 1 - 82% = 1 - 0.82 = 0.18
/ 2 = 0.18 / 2 = 0.09
Z/2 = Z 0.09 = 1.341
Margin of error = E = Z/2* (/n)
= 1.341 * (2.2/ 100 )
= 0.29
At 82% confidence interval estimate of the population mean is,
- E < < + E
55.3 - 0.29 < < 55.3 + 0.29
55.01 < < 55.59
(55.01 , 55.59)
e ) Given that,
standard deviation = = 2.2
margin of error = E = ( width / 2 = 1 / 2 ) =0.5
At 99% confidence level the z is ,
= 1 - 99% = 1 - 0.99 = 0.01
/ 2 = 0.01 / 2 = 0.005
Z/2 = Z0.005 = 2.576
Sample size = n = ((Z/2 * ) / E)2
= ((2.576 * 2.2) / 0.5)2
= 128
Sample size = 128
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