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In a sample of 1234 checked items, 20 checked items areovercharged. It is claimed that 1% of the items sold are overcharged.

The null hypothesis is written as follows.

The proportion of overcharged items is equal to 1%.

\({H_0}:p = 0.01\).

The alternative hypothesis is written as follows.

The proportion of overcharged items is not equal to 1%.

\[{H_1}:p \ne 0.01\].

The test is two-tailed.

The sample size is equal to n=1234.

The sample proportion of overcharged items is computed below.

\[\begin{array}{c}\hat p = \frac{{{\rm{Number}}\;{\rm{of}}\;{\rm{items}}\;{\rm{that}}\;{\rm{were}}\;{\rm{overcharges}}}}{{{\rm{Sample}}\;{\rm{Size}}}}\\ = \frac{{20}}{{1234}}\\ = 0.016\end{array}\].

The population proportion of overcharged items is equal to 0.016.

The value of the test statistic is computed below.

\(\begin{array}{c}z = \frac{{\hat p - p}}{{\sqrt {\frac{{pq}}{n}} }}\\ = \frac{{0.016 - 0.01}}{{\sqrt {\frac{{0.01\left( {1 - 0.01} \right)}}{{1234}}} }}\\ = 2.118\end{array}\).

Thus, z=2.118.

Referring to the standard normal table, the critical value of z at\(\alpha = 0.05\)for a two-tailed test is equal to 1.96.

Referring to the standard normal table, the p-value for the test statistic value of 2.118 is equal to 0.0342.

Asthe p-value is less than 0.05, the null hypothesis is rejected.

There is enough evidence to reject the claim that the proportion of sales that are overcharged is equal to 1%.

It can be seen that the sample proportion of overcharged items isgreater than 1% (1.6%). Thus, it can be said that the scanners are working efficiently and helping in identifying items that are overcharged.

Thus, the scanners help the customers to avoid overcharges.