Question

Testing Claims About Proportions. In Exercises 9–32, test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, or critical value(s), then state the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim. Use the P-value method unless your instructor specifies otherwise. Use the normal distribution as an approximation to the binomial distribution, as described in Part 1 of this section.

Super Bowl Wins Through the sample of the first 49 Super Bowls, 28 of them were won by teams in the National Football Conference (NFC). Use a 0.05 significance level to test the claim that the probability of an NFC team Super Bowl win is greater than one-half.

Short answer

Null hypothesis: The proportion of wins by an NFC team in the Super Bowl is equal to one-half.

Alternative hypothesis: The proportion of wins by an NFC team in the Super Bowl is greater than one-half.

Test statistic: 1.00

Critical value: 1.645

P-value: 0.1587

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 one-half.

Step-by-step solution

Given information

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

Hypotheses

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.

Sample size, sample proportion,and population proportion

The sample size is equal to n=49.

The sample proportion of overcharged items is computed below.

\[\begin{array}{c}\hat p = \frac{{{\rm{Number}}\;{\rm{of}}\;{\rm{wins}}\;{\rm{by}}\;{\rm{an}}\;{\rm{NFC}}\;{\rm{team}}}}{{{\rm{Sample}}\;{\rm{Size}}}}\\ = \frac{{28}}{{49}}\\ = 0.571\end{array}\].

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

Test statistic

The value of the test statistic is computed below.

\(\begin{array}{c}z = \frac{{\hat p - p}}{{\sqrt {\frac{{pq}}{n}} }}\\ = \frac{{0.571 - 0.5}}{{\sqrt {\frac{{0.5\left( {1 - 0.5} \right)}}{{49}}} }}\\ = 0.994\\ \approx 1.00\end{array}\).

Thus, z=1.00.

Critical value and p-value

Referring to the standard normal table, the critical value of z at\(\alpha = 0.05\)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.

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

Conclusion of the test

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%.