Question

Tests made on the breaking strength of 10 pieces of a metal gave the
following results: 578, 572, 570, 568, 572, 570, 570, 572, 598 and 584
kg. Test if the mean breaking strength of the wire can be assumed as
577 kg.

Accepted Answer

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 (H0:μ=577 Alternative Hypothesis(H1μ≠577 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  Step 4 :- Level of significance ∝ Step 5 :- Critical value The value of t∝ 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∝2.25 the null hypothesis is accepted. The mean is 577

Question #325539

Expert's answer

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