Question

Your 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 as small as the proportion in your sample or smaller? Follow the four-step process.

(b) A critic says, “Aha! You claim $90\%$, but in your sample, the on-time percentage is lower than that. So the $90\%$claim is wrong.” Explain in simple language why your probability calculation in (a) shows that the result of the sample does not refute the $90\%$claim.

Short answer

(a) $P(\hat{p}\le 0.86)=0.0918$

(b) Since the probability is greater than $0.05$, it is likely to obtain a sample with sample proportion of $0.86$if the true population proportion is $0.90$or $90\%$. Thus the claim could be correct.

Step-by-step solution

Part(a) Step 1: Given Information

Given

$$\begin{gathered} p=90\% \\ =0.90 \end{gathered}$$

$x=86$

$n=100$

Part(a) Step 2: Explanation

The sample proportion is the number of successes divided by the sample size:

$$\begin{gathered} \hat{p}=\frac{x}{n} \\ =\frac{86}{100} \\ =0.86 \end{gathered}$$

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

localid="1650094798455" $$\begin{gathered} {\mu }_{\hat{p}}=p \\ =0.90 \end{gathered}$$

The standard deviation of the sampling distribution of $\hat{p}$is:

localid="1650094821910" $$\begin{gathered} {\sigma }_{\hat{p}}=\sqrt{\frac{p(1-p)}{n}} \\ =\sqrt{\frac{0.90(1-0.90)}{100}} \\ =0.03 \end{gathered}$$

The z-score is the value decreased by the mean, divided by the standard deviation:

localid="1650094847207" $$\begin{gathered} z=\frac{x-\mu }{\sigma } \\ =\frac{0.86-0.90}{0.03} \\ \approx -1.33 \end{gathered}$$

Determine the corresponding probability using table A:

localid="1650094869728" $$\begin{gathered} P(\hat{p}\le 0.86)=P(z<-1.33) \\ =0.0918 \end{gathered}$$

Part(b) Step 1: Given Information

Given

$$\begin{gathered} p=90\% \\ =0.90 \end{gathered}$$

$x=86$

$n=100$

Part(b) Step 2: Explanation

The sample proportion is the number of successes divided by the sample size:

$$\begin{gathered} \hat{p}=\frac{x}{n} \\ =\frac{86}{100} \\ =0.86 \end{gathered}$$

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

localid="1650094921335" $$\begin{gathered} {\mu }_{\hat{p}}=p \\ =0.90 \end{gathered}$$

The standard deviation of the sampling distribution of $\hat{p}$is:

localid="1650094942237" $$\begin{gathered} {\sigma }_{\hat{p}}=\sqrt{\frac{p(1-p)}{n}} \\ =\sqrt{\frac{0.90(1-0.90)}{100}} \\ =0.03 \end{gathered}$$

The z-score is the value decreased by the mean, divided by the standard deviation:

localid="1650094966222" $$\begin{gathered} z=\frac{x-\mu }{\sigma } \\ =\frac{0.86-0.90}{0.03} \\ \approx -1.33 \end{gathered}$$

Determine the corresponding probability using table A:

localid="1650094986566" $$\begin{gathered} P(\hat{p}\le 0.86)=P(z<-1.33) \\ =0.0918 \end{gathered}$$