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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