In a study of the accuracy of fast food drive-through orders, one restaurant had 40 orders that were not accurate among 307 orders observed. Use a 0.05 significance level to test the claim that the rate of inaccurate orders is greater than 10%. State the test result in terms of the claim. Identify the null and alternative hypotheses for this test The test statistic for this hypothesis test is? The P-value for this hypothesis test is? Identify the conclusion for this hypothesis test. State the test result in terms of the claim.

We are given that in a study of the accuracy of fast food drive-through orders, one restaurant had 40 orders that were not accurate among 307 orders observed.

Let p = population proportion rate of inaccurate orders

So, Null Hypothesis, [tex]H_0[/tex] : p [tex]\leq[/tex] 10% {means that the rate of inaccurate orders is less than or equal to 10%}

Alternate Hypothesis, [tex]H_A[/tex] : p > 10% {means that the rate of inaccurate orders is greater than 10%}

The test statistics that will be used here is One-sample z-test for proportions;

T.S. = [tex]\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n} } }[/tex] ~ N(0,1)

where, [tex]\hat p[/tex] = sample proportion of inaccurate orders = [tex]\frac{40}{307}[/tex] = 0.13

n = sample of orders = 307

So, the test statistics = [tex]\frac{0.13-0.10}{\sqrt{\frac{0.10(1-0.10)}{307} } }[/tex]

= 1.75

The value of z-test statistics is 1.75.

Also, the P-value of the test statistics is given by;

P-value = P(Z > 1.75) = 1 - P(Z [tex]\leq[/tex] 1.75)

= 1 - 0.95994 = 0.04006

Now, at 0.05 level of significance, the z table gives a critical value of 1.645 for the right-tailed test.

Since the value of our test statistics is more than the critical value of z as 1.75 > 1.645, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region.

Therefore, we conclude that the rate of inaccurate orders is greater than 10%.