Question

On-time shipping A mail-order company advertises that it ships $90\%$ of its orders within three working days. You select an SRS of $100$ of the $5000$ orders received in the past week for an audit. The audit reveals that 86 of these orders were shipped on time.

a. If the company really ships $90\%$ of its orders on time, what is the probability that the proportion in an SRS of $100$ orders is $0.86$ or less?

b. Based on your answer to part (a), is there convincing evidence that less than $90\%$ of all orders from this company are shipped within three working days? Explain your reasoning.

Short answer

Part (a) $9.18\%$

Part (b) No.

Step-by-step solution

Part (a) Step 1: Given information

$$\begin{gathered} p=90\%=0.40 \\ \hat{p}=0.86 \\ n=100 \end{gathered}$$

Part (a) Step 2: Concept

$$\begin{gathered} {\sigma }_{\hat{p}}=\sqrt{\frac{p(1-p)}{n}} \\ z=\frac{x-\mu }{\sigma } \end{gathered}$$

Part (a) Step 3: Calculation

The sampling distribution of the sample proportions p has a mean of

${\mu }_{\hat{p}}=p=0.90$

The sampling distribution of the sample proportion p has a standard deviation of

$$\begin{gathered} {\sigma }_{\hat{p}}=\sqrt{\frac{p(1-p)}{n}} \\ =\sqrt{\frac{0.90(1-0.90)}{100}} \\ =0.03 \end{gathered}$$

The $z-$score is

$$\begin{gathered} z=\frac{x-\mu }{\sigma } \\ =\frac{0.86-0.90}{0.03} \\ =-1.33 \end{gathered}$$

The associating probability using the normal probability table in $P(Z<-1.33)$ is given in the standard normal probability table in the row starting with $-1.3$ and the column starting with $0.03$

$$\begin{gathered} P(\hat{p}>0.86)=P(z>-1.33) \\ =0.0918 \\ =9.18\% \end{gathered}$$

Part (b) Step 1: Calculation

The population proportion $p$ is equal to the mean of the sampling distribution of the sample proportions $\hat{p}$

${\mu }_{\hat{p}}=p=0.90$

The $10\%$ requirement states that the sample size must be less than $10\%$ of the total population size. The ten percent requirement is met because the abundance of $100$ orders is less than $10\%$ of the population of all orders in the previous week (there are $5000$ orders in the previous week).

${\sigma }_{\hat{p}}=\sqrt{\frac{p(1-p)}{n}}$

$$\begin{gathered} =\sqrt{\frac{0.90(1-0.90)}{100}} \\ =0.03 \end{gathered}$$

The square root of the product of $p$ and $(1-p)$ divided by the sample sine $n$ is the standard deviation of the sampling distribution of the sample proportion $p$

The $z-$score is

$$\begin{gathered} z=\frac{x-\mu }{\sigma } \\ =\frac{0.86-0.90}{0.03} \\ =-1.33 \end{gathered}$$

In the conventional normal probability table, in the row beginning with $-1.3$and in the column beginning with$.03$ the associating probability is given using the normal probability table $P(Z<-1.33)$

$$\begin{gathered} P(\hat{P}>0.86)=P(z>-1.33) \\ =0.0918 \\ =9.18\% \end{gathered}$$

Because the probability is not insignificant, the event is likely to happen by chance, and there is no solid proof that less than $90$ percent of all purchases from this company are dispatched within three working days.