Question

Assume that the helium porosity (in percentage) of coal samples taken from any particular seam is normally distributed with true standard deviation 0.74 (a Compute a 95% CI or the true average porosity or a certain seam f the avera e porost two decimal places.) rom, i e seam vas 48. specimens na oranswers or c 4.53 5.17 (b) Compute a 98% CI for true average porosity of another seam based on 13 specimens with a sample average porosity of 4.56. (Round your answers to two decimal places.) 4.08 , 5.04 c) How large a sample size is necessary if the width of the 95% interval is to be 0377(Round your answer up to the nearest whole number.) Xspecimens with 99% confidence? (Round your d what sample size is necessary to estimate true average porosity to within 0.23 37 e ele t ais eu o um specimens

Answer 1

Solution :

A ) Given that,

$\bar x$ = 4.85

$\sigma$ = 0.74

n = 21

At 95% confidence level the z is ,

$\alpha$  = 1 - 95% = 1 - 0.95 = 0.05

$\alpha$ / 2 = 0.05 / 2 = 0.025

Z_$\alpha$/2 = Z_0.025 = 1.960

Margin of error = E = Z_$\alpha$/2* ($\sigma$$\sqrt$/n)

= 1.960 * (0.74 / $\sqrt$ 21 )

= 0.32

At 95% confidence interval estimate of the population mean is,

$\bar x$ - E < $\mu$ < $\bar x$ + E

4.85 - 0.32 < $\mu$ < 4.85 + 0.32

4.53 < $\mu$ < 5.17

(4.53 , 5.17)

B ) Given that,

$\bar x$ = 4.56

$\sigma$ = 0.74

n = 13

At 98% confidence level the z is ,

$\alpha$  = 1 - 98% = 1 - 0.98 = 0.02

$\alpha$ / 2 = 0.02 / 2 = 0.01

Z_$\alpha$/2 = Z_0.01 =2.326

Margin of error = E = Z_$\alpha$/2* ($\sigma$$\sqrt$/n)

= 2.326 * (0.74 / $\sqrt$ 13 )

= 0.48

At 98% confidence interval estimate of the population mean is,

$\bar x$ - E < $\mu$ < $\bar x$ + E

4.56 - 0.48 < $\mu$ < 4.56 + 0.48

4.08 < $\mu$ < 5.04

(4.08 , 5.04)

C ) Given that,

$\hat p$$\hatp$ = 0.5

1 - $\hat p$ = 1 - 0.5 = 0.5

margin of error = E = 0.37 / 2 = 0.185

At 95% confidence level the z is ,

$\alpha$ = 1 - 95% = 1 - 0.95 = 0.05

$\alpha$ / 2 = 0.05 / 2 = 0.025

Z_$\alpha$/2 = Z_0.025 = 1.960

Sample size = n = ((Z_{$\alpha$ / 2}) / E)² * $\hat p$ * (1 - $\hat p$ )

= (1.960 / 0.185)² * 0.5 * 0.5

= 28.05

= 28

n = sample size = 28

D ) Given that,

$\hat p$$\hatp$ = 0.5

1 - $\hat p$ = 1 - 0.5 = 0.5

margin of error = E = 0.23

At 99% confidence level the z is ,

$\alpha$  = 1 - 99% = 1 - 0.99 = 0.01

$\alpha$ / 2 = 0.01 / 2 = 0.005

Z_$\alpha$/2 = Z_0.005 = 2.576

Sample size = n = ((Z_{$\alpha$ / 2}) / E)² * $\hat p$ * (1 - $\hat p$ )

= (2.576 / 0.23 )² * 0.5 * 0.5

= 31.36

= 31

n = sample size = 31