Question

A fertilizer mixing machine is set to give 14 kg. of nitrate for every quintal bag of fertilizer. Ten

100 kg bags are examined. The percentage of nitrate ate as follows: 11, 14, 13, 12, 15, 16, 11, 10, 19, 13

Is there reason to believe that the machine is defective?​

Answer 1

SOLUTION

GIVEN

A fertilizer mixing machine is set to give 14 kg. of nitrate for every quintal bag of fertilizer. Ten 100 kg bags are examined. The percentage of nitrate ate as follows: 11, 14, 13, 12, 15, 16, 11, 10, 19, 13

TO CHECK

Is there reason to believe that the machine is defective

EVALUATION

Here it is given that the observations are

11, 14, 13, 12, 15, 16, 11, 10, 19, 13

Mean

$\sf{ = \bar{x}}$

$\displaystyle \sf{ = \frac{Sum \: of \: the \:observations }{Total \: number \: of \: observations } }$

$\displaystyle \sf{ = \frac{134}{10} }$

$\displaystyle \sf{ =13.4}$

Standard deviation = SD

$\displaystyle \sf{ = \frac{1}{n - 1} \sum \: {(x - \bar{x})}^{2} }$

$\sf{ = 2.71}$

We test the null hypothesis $\sf{H_0 : \mu = 14}$

against the alternative hypothesis $\sf{H_1 : \mu \ne 14}$

at 1% level of significance

S.E of mean

$\displaystyle \sf{ = \frac{ \sigma}{ \sqrt{n} } }$

$\displaystyle \sf{ = \frac{ 2.71}{ \sqrt{10} } }$

$\displaystyle \sf{ =0.856}$

Now

$\displaystyle \sf{ z= \frac{ \bar{x} - \mu }{ S. E \: \: of \: mean} }$

$\displaystyle \sf{ \implies \: z= \frac{13.4 - 14 }{ 0.856} }$

$\displaystyle \sf{ \implies \: z= -0.7}$

$\displaystyle \sf{ \therefore \: \: |z| \: = 0.7 < 2.58}$

Since z does not exceeds 2.58

Thus there is no reason to reject null hypothesis

So the machine is not defective at 1% level of significance

Also it can be noted that z does not exceeds 1.64

Thus there is no reason to reject null hypothesis

So the machine is not defective at 5% level of significance

FINAL ANSWER

There is no reason to believe that the machine is defective

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