Question

1. an online retailer claims that 90% of all orders are shipped within 12 hours of being received. a consumer group placed 121 orders of different sizes and at different times of day; 102 orders were shipped within 12 hours. a. describe the sampling distribution of the sample proportion $hat{p}$. compute the standard deviation rounded to 4 decimals. $p = .90$ $n = 120$ $q = .1$ $hat{p}=sqrt{\frac{pq}{n}}=sqrt{\frac{(.90)(.1)}{120}}=.0274$ b. compute the sample proportion of items shipped within 12 hours. c. find the probability that in a sample of 121 orders, at most 102 orders were shipped within 12 hours. d. based on the answer to part (c), draw a conclusion about the retailers claim.

Explanation:

Step1: Recall sampling - distribution properties

For a sample proportion $\hat{p}$, if $np\geq5$ and $nq\geq5$, the sampling - distribution of $\hat{p}$ is approximately normal with mean $\mu_{\hat{p}} = p$ and standard deviation $\sigma_{\hat{p}}=\sqrt{\frac{pq}{n}}$, where $p$ is the population proportion, $q = 1 - p$, and $n$ is the sample size. Given $p = 0.90$, $q=1 - 0.90 = 0.10$, and $n = 121$.
$\sigma_{\hat{p}}=\sqrt{\frac{0.90\times0.10}{121}}=\sqrt{\frac{0.09}{121}}\approx0.0274$

Step2: Compute sample proportion

The sample proportion $\hat{p}$ is calculated as $\hat{p}=\frac{x}{n}$, where $x$ is the number of successes in the sample and $n$ is the sample size. Here, $x = 102$ and $n = 121$. So, $\hat{p}=\frac{102}{121}\approx0.8429$

Step3: Standardize for normal - distribution

First, find the mean of the sampling - distribution $\mu_{\hat{p}}=p = 0.90$ and the standard deviation $\sigma_{\hat{p}}\approx0.0274$. We want to find $P(\hat{p}\leq\frac{102}{121})\approx P(\hat{p}\leq0.8429)$. We standardize $\hat{p}$ using the formula $z=\frac{\hat{p}-\mu_{\hat{p}}}{\sigma_{\hat{p}}}$.
$z=\frac{0.8429 - 0.90}{0.0274}=\frac{- 0.0571}{0.0274}\approx - 2.08$
Then, $P(\hat{p}\leq0.8429)=P(Z\leq - 2.08)$. Looking up in the standard - normal table, $P(Z\leq - 2.08)=0.0188$

Step4: Draw conclusion

The probability that at most 102 out of 121 orders are shipped within 12 hours is approximately $0.0188$. Since this probability is very low (less than a common significance level of 0.05), we have evidence to doubt the retailer's claim that 90% of all orders are shipped within 12 hours.

Answer:

a. The sampling - distribution of $\hat{p}$ is approximately normal with mean $\mu_{\hat{p}} = 0.90$ and standard deviation $\sigma_{\hat{p}}\approx0.0274$.
b. $\hat{p}\approx0.8429$
c. $P(\hat{p}\leq0.8429)\approx0.0188$
d. We have evidence to doubt the retailer's claim that 90% of all orders are shipped within 12 hours.