Question

2. -/13 points DevoreStat9 7..005. Assume that the helium porosity (in percentage) of coal samples taken from any particular seam is normally distributed with true standard deviation 0.73. (a) Compute a 95% CI for the true average porosity of a certain seam if the average porosity for 22 specimens from the seam was 4.85. (Round your answers to two decimal places.) (b) Compute a 98% CI for true average porosity of another seam based on 15 specimens with a sample average porosity of 4.56. (Round your answers to two decimal places.) (c) How large a sample size is necessary if the width of the 95% interval is to be 0.36? (Round your answer up to the nearest whole number.) specimens (d) What sample size is necessary to estimate true average porosity to within 0.25 with 99% confidence? (Round your answer up to the nearest whole number.) specimens You may need to use the appropriate table in the Appendix of Tables to answer this question, Need Help? Read It Talk to a Tutor Submit Answer Practice Another Version

Answer 1

Solution :

Given that,

a) Point estimate = sample mean = $\bar x$ = 4.85

Population standard deviation = $\sigma$ = 0.73

Sample size = n = 22

At 95% confidence level the z is,

$\alpha$ = 1 - 95%

$\alpha$ = 1 - 0.95 = 0.05

$\alpha$/2  = 0.025

Z_$\alpha$/2 = Z _0.025 = 1.96

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

= 1.96 * ( 0.73 / $\sqrt$22 )

= 0.31

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

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

4.85 - 0.31 < $\mu$ < 4.85 + 0.31

4.54 < $\mu$ < 5.16

( 4.54, 5.16 )

b) Point estimate = sample mean = $\bar x$ = 4.56

Population standard deviation = $\sigma$ = 0.73

Sample size = n = 15

At 98% confidence level the z is,

$\alpha$ = 1 - 98%

$\alpha$ = 1 - 0.98 = 0.02

$\alpha$/2  = 0.01

Z_$\alpha$/2 = Z _0.01 = 2.326

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

= 2.326 * ( 0.73 / $\sqrt$15)

= 0.44

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

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

4.56 - 0.44 < $\mu$ < 4.56 + 0.44

4.12 < $\mu$ < 5.00

( 4.12, 5.00)

c) Margin of error = E = 0.36 / 2 = 0.18

At 95% confidence level the z is,

$\alpha$ = 1 - 95%

$\alpha$ = 1 - 0.95 = 0.05

$\alpha$/2  = 0.025

Z_$\alpha$/2 = Z _0.025 = 1.96

sample size = n = [Z_$\alpha$/2* $\sigma$ / E] ²

n = [ 1.96 * 0.73 / 0.18 ]²

n = 63.18

Sample size = n = 64 specimens.

d) Margin of error = E = 0.25

At 99% confidence level the z is,

$\alpha$ = 1 - 99%

$\alpha$ = 1 - 0.99 = 0.01

$\alpha$/2  = 0.005

Z_$\alpha$/2 = Z _0.005 = 2.576

sample size = n = [Z_$\alpha$/2* $\sigma$ / E] ²

n = [ 2.576 * 0.73 / 0.25 ]²

n = 56.57

Sample size = n = 57 specimens.

Answer 2

a.- 4.52-5.17

b.- 4.12- 5.0

c.-55

d.-374