Answer: Yes , the accuracy rate appear to be​ acceptable .
Step-by-step explanation:
Let p be the population proportion of the orders that were not accurate .
Then according to the claim we have ,
[tex]H_0:p=0.10\\\\ H_a:p\neq0.10[/tex]
Since the alternative hypothesis is two-tailed so the hypothesis test is a  two-tailed test.
For sample ,
n = 391
Proportion of  the orders that were not accurate =[tex]\hat{p}=\dfrac{34}{391}\approx0.087[/tex]
Test statistics for population proportion :-
[tex]z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}\\\\=\dfrac{0.087-0.10}{\sqrt{\dfrac{0.10(0.90)}{391}}}\approx-0.86[/tex]
By using the standard normal distribution table,
The p-value : [tex]2(z>-0.86)=0.389789\approx0.39[/tex]
Since the p-value is greater that the significance level (0.05), so we do not reject the null hypothesis.
Hence, we conclude that the accuracy rate appear to be​ acceptable.
Assuming the rate of inaccurate orders is equal to​ 10%. The accuracy rate appear to be​ acceptable.
H o rate = 10%
H a rate is not equal to 10%
alpha=0.05
Test statistic is a one sample proportion z,
Where:
z=( phat- p)/√((p ×(1- p)/n))
phat=34/391
phat=0.0869
z=-(0.0869-0.05)/√(.10×.90/391)
z=-0.0369/√(.10×.90/391)
z=-.0369/0.01517
z=-2.43
The accuracy rate appear to be​ acceptable.
Inconclusion the accuracy rate appear to be​ acceptable.
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