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Nine items of a sample had the following values 45 ,47 ,50 ,48 ,47 ,49 ,53 ,51.Does the mean of 9 items differ significantly from the assumed population mean 47.5 ?
$\text{First we calculate$\bar{X}$and s$^2$.}$
$\text{$\bar{X}$=$\frac{(45+47+50+48+47+49+53+51+52)}{9}$=49.11}$ \begin{array}{|c|c|c|c|c|c|c|c|c|c|c|} \hline X & 45 & 47 & 50 &52 &48 &47 &49 &53 &51 &SUM \\ \hline X - \bar{X} &-4.11 &-2.11 &0.89 &2.89 &-1.11 &-2.11 &-0.11 &3.89 &1.89 \\ \hline (X – \bar{X})^2 &16.89 &4.45 &0.79 &8.35 &1.23 &4.45 &1.21 &15.13 &3.57 &56.07 \\ \hline \end{array} $\text{s$^2$= ∑$\frac{(X – \bar{X})^2 }{9}$=$\frac{56.07}{9}$= 6.23}$ $\text{(i) The null hypothesis H$_0$: μ = 47.5. Alternative hypothesis H$_1$: μ ≠ 47.5}$ $\text{(ii) Calculation of test statistic: Since the sample size is small , we use t – distribution;}$ $\text{t =$\frac{(\bar{X}-μ)}{(\frac{s}{\sqrt{(n-1)}})}$=$\frac{(49.11-47.5)}{\frac{( \sqrt{6.23}}{\sqrt{(9-1)}})}$= 1.82}$ $\text{(iii) Level of significance : α =0.05}$ $\text{(iv) Critical value : the value of t$_α$at 5% level of significance for ν= 9-1 =8 degrees of freedom is 2.306}$ $\text{(v) Decision : since the calculated value of |t| =1.82 is less than the table value t$_α$=2.306. Therefore, the null hypothesis is accepted.}$