Complete Question
In a study of the accuracy of fast food drive-through orders, one restaurant had 32 orders that were not accurate among 367 orders observed. Use a 0.05 significance level to test the claim that the rate of inaccurate orders is equal to 10%. Does the accuracy rate appear to be acceptable?
Answer:
The decision rule  is
 Fail to reject the null hypothesis
The conclusion is Â
 There is sufficient evidence to show that the rate of inaccurate orders is equal to​ 10%
Step-by-step explanation:
Generally from the question we are told that
  The sample size is  n =  367
  The number of orders that were not accurate is  [tex]k = 32[/tex]
  The population proportion for rate of inaccurate orders is  p = 0.10
The null hypothesis is  [tex]H_o : p = 0.10[/tex]
The alternative hypothesis is  [tex]H_a : p \ne 0.10[/tex]
Generally the sample proportion is mathematically represented as Â
     [tex]\^ p = \frac{k}{n}[/tex]
=> Â Â Â [tex]\^ p = \frac{32}{367}[/tex]
=> Â Â Â [tex]\^ p = 0.0872[/tex]
Generally the test statistics is mathematically represented as
     [tex]z= \frac{ \^ p - p }{ \sqrt{ \frac{ p(1 - p)}{ n} } }[/tex]
=> Â Â Â [tex]z= \frac{ 0.0872 - 0.10 }{ \sqrt{ \frac{ 0.10 (1 - 0.10 )}{367} } }[/tex] Â
=> Â Â Â [tex]z= -0.8174[/tex]
From the z table  the area under the normal curve to the left corresponding to  -0.8174 is Â
     [tex]P(z < -0.8173 ) = 0.20688[/tex]
Generally the p-value is mathematically represented as
     [tex]p- value = 2 * 0.20688[/tex]
=> Â Â Â [tex]p- value = 0.4138[/tex]
From the value obtained we see that  [tex]p-value > \alpha[/tex] hence
The decision rule  is
 Fail to reject the null hypothesis
The conclusion is Â
 There is sufficient evidence to show that the rate of inaccurate orders is equal to​ 10%
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