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.

Touch Therapy Repeat the preceding exercise using a 0.01 significance level. Does the conclusion change?

Out of 280 trials, 123 guesses were correct by the touch therapists. It is claimed that that touch therapists randomly guess the answer.

The null hypothesis is written as follows.

The proportion of correct guesses is equal to 0.5.

\({H_0}:p = 0.5\).

The alternative hypothesis is written as follows.

The proportion of correct guesses is not equal to 0.5.

\({H_1}:p \ne 0.5\).

The test is two-tailed.

The sample size is n=280.

The sample proportion of correct guesses is computed below.

\[\begin{array}{c}\hat p = \frac{{{\rm{Number}}\;{\rm{of}}\;{\rm{correct}}\;{\rm{guesses}}}}{{{\rm{Sample}}\;{\rm{Size}}}}\\ = \frac{{123}}{{280}}\\ = 0.439\end{array}\].

The population proportion of correct guesses is equal to 0.5.

The value of the test statistic is computed below.

\(\begin{array}{c}z = \frac{{\hat p - p}}{{\sqrt {\frac{{pq}}{n}} }}\\ = \frac{{0.439 - 0.5}}{{\sqrt {\frac{{0.5\left( {1 - 0.5} \right)}}{{280}}} }}\\ = - 2.032\end{array}\).

Thus, z=-2.032.

Referring to the standard normal table, the critical value of z at\(\alpha = 0.01\)for a two-tailed test is equal to 2.5758.

Referring to the standard normal table, the p-value for the test statistic value of -2.032 is equal to 0.0422.

Asthe p-value is greaterthan 0.01, the null hypothesis is failed to reject.

There is not enough evidence to reject the claim that the touch therapists randomly guess the correct answer.