91影视

Out of 879 challenges made by single players, a total of 231 areoverturned. It is claimed that less than \(\frac{1}{3}\)of challenges are successful.

The null hypothesis is written as follows.

The proportion of successful challenges is equal to\(\frac{1}{3}\).

\({H_0}:p = 0.333\).

The alternative hypothesis is written as follows.

The proportion of successful challenges is less than\(\frac{1}{3}\).

\({H_1}:p < 0.333\).

The test is left-tailed.

The sample size is equal to n=879.

The sample proportion of successful challenges is computed below.

\(\begin{array}{c}\hat p = \frac{{{\rm{Number}}\;{\rm{of}}\;{\rm{successful}}\;{\rm{challenges}}}}{{{\rm{Sample}}\;{\rm{Size}}}}\\ = \frac{{231}}{{879}}\\ = 0.263\end{array}\).

The population proportion of successful challenges is equal to 0.333.

The value of the test statistic is computed below.

\(\begin{array}{c}z = \frac{{\hat p - p}}{{\sqrt {\frac{{pq}}{n}} }}\\ = \frac{{0.263 - 0.333}}{{\sqrt {\frac{{0.333\left( {1 - 0.333} \right)}}{{879}}} }}\\ = - 4.404\end{array}\).

Thus, z=-4.404.

Referring to the standard normal table, the critical value of z at\(\alpha = 0.01\)for a left-tailed test is equal to -2.3263.

Referring to the standard normal table, the p-value for the test statistic value of -4.404 is equal to 0.00003.

Asthe p-value is less than 0.01, the null hypothesis is rejected.

There is enough evidence to support the claim that the proportion of successful challenges is less than\(\frac{{\bf{1}}}{{\bf{3}}}\).

About 26.3% of the challenges raised areoverturned, which doesn鈥檛 seem to be a very high percentage. Thus, the ability of referees to make the correct decision is evident.