In a study of the accuracy of fast food drive-through orders, McDonald’s had 33 orders that were not accurate among 362 orders observed. Use

Question

In a study of the accuracy of fast food drive-through orders, McDonald’s had 33 orders that were not accurate among 362 orders observed. Use a 0.05 significance level to test the claim that the proportion of inaccurate orders is equal to 10%. A: Write the claim as a mathematical statement involving the population proportion p. B: State the null (H0) and alternative (H????) hypotheses. C: Find the test statistic (z????????????????) for the given sample. D: Find the critical value(s) OR the P-value for the given test. E: Would you Reject or Fail to Reject the null (H0) hypothesis. F: Write the conclusion of the test.

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Jade 2 weeks 2021-09-13T13:39:16+00:00 1 Answer 0

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    2021-09-13T13:40:22+00:00

    Answer:

    A. We need to conduct a hypothesis in order to test the claim that the true proportion of inaccurate orders p is 0.1.

    B. Null hypothesis:p=0.1  

    Alternative hypothesis:p \neq 0.1  

    C. z=\frac{0.0912 -0.1}{\sqrt{\frac{0.1(1-0.1)}{362}}}=-0.558  

    D. z_{\alpha/2}=-1.96  z_{1-\alpha/2}=1.96

    E. Fail to the reject the null hypothesis

    F. So the p value obtained was a very high value and using the significance level given \alpha=0.05 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the true proportion of inaccurate orders is not significantly different from 0.1.  

    Step-by-step explanation:

    Data given and notation

    n=362 represent the random sample taken

    X=33 represent the number of orders not accurate

    \hat p=\frac{33}{363}=0.0912 estimated proportion of orders not accurate

    p_o=0.10 is the value that we want to test

    \alpha=0.05 represent the significance level

    Confidence=95% or 0.95

    z would represent the statistic (variable of interest)

    p_v represent the p value (variable of interest)  

    A: Write the claim as a mathematical statement involving the population proportion p

    We need to conduct a hypothesis in order to test the claim that the true proportion of inaccurate orders p is 0.1.

    B: State the null (H0) and alternative (H1) hypotheses

    Null hypothesis:p=0.1  

    Alternative hypothesis:p \neq 0.1  

    When we conduct a proportion test we need to use the z statistic, and the is given by:  

    z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

    The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

    C: Find the test statistic

    Since we have all the info required we can replace in formula (1) like this:  

    z=\frac{0.0912 -0.1}{\sqrt{\frac{0.1(1-0.1)}{362}}}=-0.558  

    D: Find the critical value(s)

    Since is a bilateral test we have two critical values. We need to look on the normal standard distribution a quantile that accumulates 0.025 of the area on each tail. And for this case we have:

    z_{\alpha/2}=-1.96  z_{1-\alpha/2}=1.96

    P value

    It’s important to refresh the p value method or p value approach . “This method is about determining “likely” or “unlikely” by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed”. Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

    The significance level provided \alpha=0.05. The next step would be calculate the p value for this test.  

    Since is a bilateral test the p value would be:  

    p_v =2*P(z<-0.558)=0.577  

    E: Would you Reject or Fail to Reject the null (H0) hypothesis.

    Fail to the reject the null hypothesis

    F: Write the conclusion of the test.

    So the p value obtained was a very high value and using the significance level given \alpha=0.05 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the true proportion of inaccurate orders is not significantly different from 0.1.  

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