Question

Homework: HW 9.1 9.2 9.3 25 of 40 (26 complete) Score: 0 of 1 pt 9.2.17 A simple random sample of size n is drawn from a population that is normally distributed. The sample mean, X, is found to be 106, and the sample standard deviation, s, is found to be 10. (a) Construct a 90% confidence interval about if the sample size, n, is 18. (b) Construct a 90% confidence interval about if the sample size, n, is 29. (c) Construct a 98% confidence interval about if the sample size, n, ls 18. (d) Could we have computed the confidence intervals in parts (a)-c) if the population had not been normally distributed? Click the loon to view the table of areas under the I-distribution. (a) Construct a 90% confidence interval about if the sample size, n, is 18. Lower bound: :Upper bound: (Use ascending order. Round to one decimal place as needed.)

Answer 1

$\mathbf{Solution}$

given :ī= 106, s = 10

(a) n = 18

degrees of freedom df) = n-1 = 18 – 1 = 17

for 90% confidence level with degrees of freedom = 17

t* = 1.740 (from t -table)

download link to calculate t*

https://drive.google.com/open?id=1z1WHfk8Oa_ZRAgh2GPyQYMO8erimrxaa

$\dpi{100} \mathbf{formula:}\; Confidence\; interval\; for\; the \;population\; mean,\;\mu$

$\dpi{100} =\bar X \pm t^{*}*\left \{ \frac{s}{\sqrt{n}} \right \}$

= 106 +1.740*1、运了

= 106 +4.1

Lower bound = 106 – 4.1 = 101.9

Upper bound = 106 + 4.1 = 110.1

(b) n=29

degrees of freedom df) = n-1 = 29 – 1 = 28

for 90% confidence level with degrees of freedom = 28

t* = 1.701 (from t -table)

download link to calculate t*

https://drive.google.com/open?id=1z1WHfk8Oa_ZRAgh2GPyQYMO8erimrxaa

$\dpi{100} \mathbf{formula:}\; Confidence\; interval\; for\; the \;population\; mean,\;\mu$

$\dpi{100} =\bar X \pm t^{*}*\left \{ \frac{s}{\sqrt{n}} \right \}$

{67/) * LOL

= 106 +3.2

Lower bound = 106 – 3.2 = 102.8

Upper bound = 106 +3.2 = 109.2

$\dpi{100} \mathbf{{\color{Blue} B.}\;{\color{Red} As \;the\; sample\; size\;increases, \;the\; margin \;of \;error\; decreases.}}$

(c) n=18

degrees of freedom df) = n-1 = 18 – 1 = 17

for 98% confidence level with degrees of freedom = 17

t* = 2.567 from t - table)

download link to calculate t*

https://drive.google.com/open?id=1z1WHfk8Oa_ZRAgh2GPyQYMO8erimrxaa

$\dpi{100} \mathbf{formula:}\; Confidence\; interval\; for\; the \;population\; mean,\;\mu$

$\dpi{100} =\bar X \pm t^{*}*\left \{ \frac{s}{\sqrt{n}} \right \}$

了10 = 106 +2.567*3、石了

= 106 +6.1

Lower bound = 106 - 6.1 = 99.9

Upper bound = 106 +6.1 = 112.1

$\dpi{100} \mathbf{{\color{Blue} B.}\;{\color{Red} As \;the \;percent\; confidence\; increases, \;the \;size \;of \;the \;interval \;increases.} }$

$\dpi{100} \mathbf{{\color{Blue} (d)}\;{\color{Blue} B.}\;{\color{Red} No,\; the \;population\; needs\; to\; be\;normally\; distributed}.}$