When she was 9 years of age, a girl 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 therapists to identify the selected hand by placing their hand just under her hand without seeing it and without touching it. Among 280 trials, the touch therapists were correct 123 times. 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? Let p denote the population proportion of correct hand selections by touch therapists. Identify the null and alternative hypotheses to test the claim that touch therapists use a method equivalent to random guesses. \( H_{0} p=0.5 \) \( H_{1}: p \neq 0.5 \) (Type integers or decimals. Do not round.) Identify the test statistic. \( z=-2.03 \) (Round to two decimal places as needed.) Identify the P-value. P-value \( =\square \) (Round to three decimal places as needed.) (

We are given that among \(280\) trials, the touch therapists correctly identified the selected hand \(123\) times. We denote by \( p \) the true proportion of correct identifications. **Step 1. Set up Hypotheses** The claim is that touch therapists’ performance is equivalent to random guessing. Since a random guess produces correct identification \(50\%\) of the time, the hypotheses are: \[ H_0: p = 0.5 \] \[ H_1: p \neq 0.5 \] **Step 2. Compute the Test Statistic** The test statistic for a proportion is given by \[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} \] where - \(\hat{p}\) is the sample proportion, - \(p_0 = 0.5\) under the null hypothesis, - \(n = 280\). First, calculate the sample proportion: \[ \hat{p} = \frac{123}{280} \approx 0.4393 \] Now, compute the standard error: \[ \text{SE} = \sqrt{\frac{0.5(1-0.5)}{280}} = \sqrt{\frac{0.25}{280}} \approx \sqrt{0.000892857} \approx 0.0299 \] Then, the test statistic is: \[ z = \frac{0.4393 - 0.5}{0.0299} \approx \frac{-0.0607}{0.0299} \approx -2.03 \] **Step 3. Determine the P-value** Since this is a two-tailed test, the P-value is: \[ \text{P-value} = 2 \times P(Z < -2.03) \] Looking up \( Z = -2.03 \) in the standard normal table, we find: \[ P(Z < -2.03) \approx 0.0212 \] Thus, the P-value is: \[ \text{P-value} \approx 2 \times 0.0212 = 0.0424 \quad \text{(rounded to three decimals: } 0.042 \text{)} \] **Final Answers:** - Null hypothesis: \( H_0: p = 0.5 \) - Alternative hypothesis: \( H_1: p \neq 0.5 \) - Test statistic: \( z = -2.03 \) - P-value: \( 0.042 \)