Question

erity w o rth wo

Answer 1

Note -that Image is not clear , last part is not visible at all .

Given

x = 34

n = 355

Thus Estimate of sample proportion of inaccuracy in order is given by

$\hat{p}$ = x/n = 34 / 355 = 0.09577465

Thus    $\hat{p}$ = 0.09577465

Null and alternative hypothesis are

B) H0 : p = 0.1

     H1 : p $\neq$ 0.1

For this hypothesis test we use one sample proportion z-test

Test Statistics TS :

TS =$\frac{\hat{p}-p}{\sqrt{p*(1-p)/n}}$

now n = 355 , p = 0.1 , $\hat{p}$ = 0.09577465

Thus

TS =$\frac{\hat{p}-p}{\sqrt{p*(1-p)/n}}$

     = $\frac{0.09577465-0.1}{\sqrt{0.1*(1-0.1)/355}}$ = -0.00422535 / 0.01592235 = -0.2653723

Thus Test Statistics for his hypothesis test is -0.2654

To compute P-value

Since alternative hypothesis is of " $\neq$ " type , so this is two-tail test .

Hence P-value is given by

P-value = P ( Z < - | TS | ) + P ( Z > | TS | )

              

here Z ~ N(0,1)

Thus P-value = P ( Z < - | TS | ) + P ( Z > | TS | )

   P-value = 2*P ( Z < - | TS | ) { due to symmetry }

    P-value = 2*P ( Z < -0.2654 )

Now P ( Z < -0.2654 ) can be computed from statistical book or more accurately from any software like R,Excel

From R

> 2*pnorm(-0.2654 ,m=0,sd=1)           # 2*P ( Z < -0.2654 )
[1] 0.7907013

thus 2*P ( Z < -0.2654 ) = 0.7907013

Hence P-Value = 0.7907013

rounding to four decimals

P-value = 0.7907

Conclusion:

Since P-Value is 0.7907 which is greater than 0.011 given significance level

i.e P-value = 0.7907 > 0.01

Hence we do not reject null hypothesis H0 at 1% of level of significance .

Correct Options is

Fail to reject H0 , there is not sufficient evidence to that reject the claim that inaccuracy in orders is equal to 10% .

{ i.e we conclude that inaccuracy in order may be equal to 10% }

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