Question

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.

Answer 1

We conclude that the rate of inaccurate orders is greater than ​10%.

Explanation:

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,
`H_0` : p
`\leq` 10% {means that the rate of inaccurate orders is less than or equal to ​10%}

Alternate Hypothesis,
`H_A` : 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. =
`\frac{\hat p-p}{\sqrt{(p(1-p))/(n) } }` ~ N(0,1)

where,
`\hat p` = sample proportion of inaccurate orders =
`(40)/(307)` = 0.13

n = sample of orders = 307

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

= 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
`\leq` 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%.