Question

1.70 9.86 5.44 4.28 4.5 8.76 9.16 6.28 3.83 3.17 5.98 2.77 3.59 3.17 8.46 7.76 5.55 5.95 9.56 3.58 p-value SDev2503

Answer 1

$(a)~95\%~C.I.~for~\mu~(=population~mean~percentage~of~SIO_2)~is\\ \left(\bar{x}-t_{0.025,19}\times \frac{s}{\sqrt{20}},~\bar{x}+t_{0.025,19}\times \frac{s}{\sqrt{20}} \right )=(4.501,6.843)\\ where,~\bar{x}=sample~mean=5.672,~s=sample~sd=2.503,~t_{0.025,19}=2.093\\ (b)~Null~hypothesis,~H_0:\mu=6.2~vs.~Alternative~hypothesis,~H_a:\mu\neq 6.2\\ t=\frac{\sqrt{22}(\bar{x}-6.2)}{s}=-0.94\\ P-value=2P(t>0.94|t\sim t_{19})=0.357>0.05.\\ Hence~we~fail~to~reject~H_0~at~5\%~level~of~significance,~hence\\ there~is~not~enough~evidence~to~conclude~that~the~mean~percentage~of~SIO_2~is\\ different~from~6.2\%.\\ (c)~From~the~plot,~we~conclude~that~assumption~of~normality~is~valid~since\\~all~points~are~very~closed~to~reference~line.$

One-Sample T: Percentage of SIO2

Test of mu = 6.2 vs not = 6.2

Variable N Mean StDev SE Mean 95% CI T P
Percentage of SIO2 20 5.672 2.503 0.560 (4.501, 6.843) -0.94 0.357