Answer
There is not sufficient evidence to reject the claim that touch therapists use a method similar to random guess.
Work Step by Step
$H_{0}:p=50$%=0.5. $H_{a}:p\ne0.5.$ $ \hat{p}$ is the number of objects with a specified value divided by the sample size. Hence $\hat{p}=\frac{x}{n}=\frac{123}{280}=0.4393.$ The test statistic is:$z=\frac{\hat{p}-p}{\sqrt{p(1-p)/n}}=\frac{0.43933-0.5}{\sqrt{0.5(1-0.5)/280}}=-2.03.$ The P is the probability of the z-score being more than 2.03 or less than -2.03 is the sum of the probability of the z-score being less than -2.03 plus 1 minus the probability of the z-score being less than 2.03, hence:P=0.0212+1-0.9788=0.0424. If the P-value is less than $\alpha$, which is the significance level, then this means the rejection of the null hypothesis. Hence:P=0.0424 is more than $\alpha=0.01$, hence we fail to reject the null hypothesis. Hence we can say that there is not sufficient evidence to reject the claim that touch therapists use a method similar to random guess.
