Question

Chicken Delight claims that 87% of its orders are delivered within 10 minutes of the time the order is placed. A sample of 80 orders revealed that 61 were delivered within the promised time. At the 0.10 significance level, can we conclude that less than 87% of the orders are delivered in less than 10 minutes?
a.What is the decision rule? (Negative amount should be indicated by a minus sign. Round your answer to 2 decimal places.)b.Compute the value of the test statistic. (Negative amount should be indicated by a minus sign. Round the intermediate values and final answer to 2 decimal places.)c.What is your decision regarding the null hypothesis?

Answer 1

Answer:

There is strong evidence that less than 87% of the orders are delivered in less than 10 minutes.

Decision rule: Reject the null hypothesis if (P-value < level of significance)

Test statistic z=-2.70

Decision: Reject the null hypothesis (0.003<0.010)

Step-by-step explanation:

In this question we have to test an hypothesis.

The null and alternative hypothesis are:

$H_0: \pi\geq0.87\\\\H_a:\pi$

The significance level is assumed to be 0.01.

The sample of size n=80 gives a proportion of p=61/80=0.7625.

The standard deviation is:

$\sigma=\sqrt{\frac{\pi(1-\pi)}{N} }=\sqrt{\frac{0.87*0.13}{80} }=0.0376$

The statistic z is then

$z=\frac{p-\pi+0.5/N}{\sigma}=\frac{0.7625-0.87+0.5/80}{0.0376} } =\frac{-10125}{0.0376} =-2.70$

The P-value is

$P(z$

The P-value (0.003) is smaller than the significance level (0.010), so the effect is significant. The null hypothesis is rejected.

There is strong evidence that less than 87% of the orders are delivered in less than 10 minutes.

Decision rule: Reject the null hypothesis if (P-value < level of significance)

Test statistic z=-2.70

Decision: Reject the null hypothesis (0.003<0.010)