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 When she was 9 years of age, Emily Rosa did a science fair experiment in which she tested professional touch therapists to see if they could sense her energy field. She flipped a coin to select either her right hand or her left hand, and then she asked the therapiststo identify the selected hand by placing their hand just under Emily’s hand without seeing it and without touching it. Among 280 trials, the touch therapists were correct 123 times (based on data in “A Close Look at Therapeutic Touch,” Journal of the American Medical Association, Vol. 279, No. 13). Use a 0.10 significance level to test the claim that touch therapists use a method equivalent to random guesses. Do the results suggest that touch therapists are effective?

Out of 280 trials, 123 guesses arecorrect 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 ofcorrect 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\(\alpha = 0.10\)for a two-tailed test is equal to 1.645.

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 less than 0.10, the null hypothesis is rejected.

There is enough evidence to reject the claim that the touch therapists randomly guess the correct answer.

Asthe sample proportion of correct guesses equal to 43.9% is even less than 50% (as proposed by the method of random guesses), it can be said that touch therapists are not at all effective.