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In a sample of 49 Super Bowls, 28 of them were won by an NFC team.

The null hypothesis is written as follows.

The proportion of wins by an NFC team is equal to one-half.

$H_0:p=0.5$

The alternative hypothesis is written as follows.

The proportion of wins by an NFC team is greater than one-half.

$H_1:p>0.5$

The test is right-tailed.

The sample size is equal to n=49.

The sample proportion of overcharged items is computed below.

$$\begin{gathered} \hat{p}=\frac{\text{Number}\text{of}\text{wins}\text{by}\text{an}\text{NFC}\text{team}}{\text{Sample}\text{Size}} \\ =\frac{28}{49} \\ =0.571 \end{gathered}$$

The population proportion of wins by an NFC team is equal to 0.5.

The value of the test statistic is computed below.

$$\begin{gathered} z=\frac{\hat{p}-p}{\sqrt{\frac{pq}{n}}} \\ =\frac{0.571-0.5}{\sqrt{\frac{0.5\left(1-0.5\right)}{49}}} \\ =0.994 \\ 鈮�1.00 \end{gathered}$$

Thus, z=1.00.

Referring to the standard normal table, the critical value of z at for a right-tailed test is equal to 1.645.

Referring to the standard normal table, the p-value for the test statistic value of 1.00 is equal to 0.1587.

As the p-value is greater than 0.05, the null hypothesis is not rejected.

There is not enough evidence to support the claim that the proportion of sales wins by an NFC team in the Super Bowl is greater than 50%.