Mumbai University > Mechanical Engineering > Sem 4 > Applied Mathematics IV

Marks: 6M

Year: Dec 2015

$$\therefore \ \overline{x}=a+\frac{\sum{d_i}}{n}=580+\left(\frac{-48}{10}\right)=575.2\therefore \sum{{\left(x_i-\overline{x}\right)}^2={\sum{d_i}}^2}-\frac{{\left(\sum{d_i}\right)}^2}{n}=912-\frac{{\left(-48\right)}^2}{10}=633.6$$ $$\therefore s^2=\frac{\sum{{\left(x_i-\overline{x}\right)}^2}}{n}=\frac{681.6}{10}=68.18$$

Step 2 :-

Null Hypothesis (H${}_{0}$)=$\mu$=577

Alternative Hypothesis(H${}_{x}$)=$\mu$$\mathrm{\neq}$70 Step 3 :- Test statistic $$t=\frac{\overline{x}-\mu }{{s}/{\sqrt{n-1}}}=\frac{575.2-577}{{\sqrt{68.16}}/{\sqrt{10-1}}}=-0.65$$ $$\mathrm{\therefore }|t| = 0.65$$

Step 4 :-

Level of significance $\mathrm{\propto }$=0.05

Step 5 :-

Critical value

The value of t${}_{\mathrm{\propto }}$ at 5% level of significance for v=10-1=9

Degree of freedom is 2.25

Step 6 :- Decision

Since the calculated value of |t|=0.65 is less than the table value t${}_{\mathrm{\propto }}$=2.25 the null hypothesis is accepted

The mean is 577