Accuracy of price scanners at Walmart. The National Institute for Standards and Technology (NIST) mandates that for every 100 items scanned through the electronic checkout scanner at a retail store, no more than 2 should have an inaccurate price. A study of the accuracy of checkout scanners at Walmart stores in California was conducted. At each of 60 randomly selected Walmart stores, 100 random items were scanned. The researchers found that 52 of the 60 stores had more than 2 items that were inaccurately priced.

a. Give an estimate of p, the proportion of Walmart stores in California that have more than 2 inaccurately priced items per 100 items scanned.

b. Construct a 95% confidence interval for p.

c. Give a practical interpretation of the interval, part b.

d. Suppose a Walmart spokesperson claims that 99% of California Walmart stores are in compliance with the NIST mandate on the accuracy of price scanners. Comment on the believability of this claim.

e. Are the conditions for a valid large-sample confidence interval for p satisfied in this application? If not, comment on the validity of the inference in part d.

f. Determine the number of Walmart stores that must be sampled in order to estimate the proper proportion to within .05 with 90% confidence using the large-sample method.a)

Let the sample size be 60, and the number of successes be 52.

Since the sample proportion's mean of the sampling distribution $\left(\hat{p}\right)$is the population proportion p. The sample proportion $\left(\hat{p}\right)$is an unbiased estimator of the population proportion $\left(\hat{p}\right)$.

Thus, the point estimate $\left(\hat{p}\right)$ of the population proportion p is obtained below:

$\hat{p}=\frac{x}{n}$

Were,

x is the number of successes in the sample

n is the sample size

Therefore,

$$\begin{gathered} \hat{p}=\frac{52}{60} \\ =0.867 \end{gathered}$$

Hence, the point estimate of the population proportion p is 0.867.

Let the confidence level is 0.95.

$$\begin{gathered} 1-\alpha =0.95 \\ \alpha =0.05 \\ \frac{\alpha }{2}=0.025 \end{gathered}$$

From the z table, the required z_0.025 value for the 95% confidence level is 1.96. Thus, $z_{0.025}=1.96$

The 95% confident interval is obtained as shown below:

$$\begin{gathered} \hat{p}\pm z_{0.025}\left(\sqrt{\frac{\hat{p}\hat{q}}{n}}\right)=0.867\pm 1.96\sqrt{\frac{\left(0.867\right)\left(0.133\right)}{60}} \\ =0.867\pm 1.96\left(\sqrt{0.001922}\right) \\ =0.867\pm 1.96\left(0.0438\right) \end{gathered}$$

$$\begin{gathered} =0.867\pm 0.0858 \\ =\left(0.867-0.0858,0.867+0.0858\right) \\ =\left(0.7812,0.9528\right) \end{gathered}$$

Thus, the 95% confidence interval for the population proportion is (0.7812,0.9528).

For 95% confidence, the proper proportion of Wal-Mart stores with more than 2 inaccurately priced items per 100 items lies between 0.7812 and 0.9528.

The 1% of California Wal-Mart stores are not in compliance with the NIST mandate on the accuracy of price scanners. That is, 0.01 of California Walmart stores are not in compliance with the NIST mandate on the accuracy of price scanners.

The value of 0.01 does not contain the 95% interval in part (b). That is, the value of 0.01 does not fall between 0.7812 and 0.9528.

Hence, the claim that “99% of California Wal-Mart stores comply with the NIST mandate on the accuracy of price scanners” is unbelievable.

The sample size is large enough if the below conditions are satisfied.

• $n\hat{p}>15q$
• $n\left(1-\hat{p}\right)⩾15$

Let,

$$\begin{gathered} n\hat{p}=60\left(0.867\right) \\ =52.02\left[⩾15\right] \end{gathered}$$

Also,

$$\begin{gathered} n\left(1-\hat{p}\right)=60\left(1-0.867\right) \\ =7.98 \\ =8\left[<15\right] \end{gathered}$$

Since the value of $n\left(1-\hat{p}\right)$is lesser than 15, the sample size is not sufficiently large. Thus, it can be concluded that the normal approximation is not reasonable.

The formula for sample size is given below:

$\mathit{n}\mathbf{=}\frac{\left(z_{\frac{\alpha }{2}}\right)\mathbf{p}\mathbf{q}}{{\left(ME\right)}^{\mathbf{2}}}$

Here, the product pq is unknown, which can be obtained by using the sample fraction of success$\hat{p}$ . Thus, in the above equation, substitute 1.645 for , 0.867 for, 0.133 for, and 0.05 for ME.

$$\begin{gathered} n=\frac{{\left(1.645\right)}^2\left(0.867\right)\left(0.133\right)}{{\left(0.05\right)}^2} \\ =\frac{\left(2.706\right)\left(0.867\right)\left(0.133\right)}{0.0025} \\ \approx 125 \end{gathered}$$

Thus, the required sample size is 125.

Hence, a sample size of 125 is necessary to estimate the true proportion to be within 0.05.