In a science fair project. Emily conducted an experiment in which she tested professional touch therapists to see if they could sense her energy field. She flipped a coin t left hand, and then she asked the therapists to identify the selected hand by placing their hand just under Emily's hand without seeing it and without touching it. Among 3 . 163 times. Complete parts (a) through (d). a. Given that Emily used a coin toss to select either her right hand or her left hand, what proportion of correct responses would be expected if the touch therapists made ra 0.5 (Type an integer or a decimal. Do not round.) b. Using Emily's sample results, what is the best poinf estimate of the therapists' success rate? 0.481 (Round to three decimal places as needed.) c. Using Emily's sample results, construct a $$ 95 \% $$ contidence interval estimate of the proportion of correct responses made by touch therapists. $$ \square

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**(a)**  
Since Emily used a fair coin toss to pick her hand, under the assumption of chance the probability of selecting the left (or right) hand is  
$$
p = 0.5.
$$

**(b)**  
Let $$\hat{p}$$ be the sample proportion of correct identifications. Given Emily’s experiment produced a success rate of 0.481, the best point estimate of the therapists’ success rate is  
$$
\hat{p}=0.481.
$$

**(c)**  
A 95% confidence interval for a proportion is given by  
$$
\hat{p} \pm z_{\alpha/2}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}},
$$  
where $$z_{\alpha/2} = 1.96$$ for a 95% confidence level and $$n$$ is the number of trials. In Emily’s experiment the number of correct responses was 163. (It turns out that with 163 successes the total number of trials works out so that $$\hat{p}=0.481$$; one may show that $$n\approx339$$ because $$163/339\approx0.481$$.)

First, compute the standard error:  
$$
SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} = \sqrt{\frac{0.481\,(1-0.481)}{339}} = \sqrt{\frac{0.481\cdot0.519}{339}}.
$$  
Perform the multiplication in the numerator:  
$$
0.481\cdot0.519 \approx 0.2493.
$$  
Then  
$$
SE \approx \sqrt{\frac{0.2493}{339}} \approx \sqrt{0.000735} \approx 0.0271.
$$

The margin of error (ME) is  
$$
ME = 1.96\times 0.0271 \approx 0.0531.
$$

Thus, the 95% confidence interval is  
$$
0.481 - 0.0531 < p < 0.481 + 0.0531,
$$  
which is approximately  
$$
0.428 < p < 0.534.
$$

**(d)**  
To decide whether the therapists performed better than chance we test:  
$$
H_0: p = 0.5 \quad \text{versus} \quad H_a: p \neq 0.5.
$$

Using the sample proportion and standard error calculated above, the test statistic is  
$$
z = \frac{\hat{p}-0.5}{SE} = \frac{0.481-0.5}{0.0271} \approx \frac{-0.019}{0.0271} \approx -0.70.
$$

A $$z$$-value of about $$-0.70$$ gives a $$p$$-value well above 0.05. In addition, note that the 95% confidence interval $$(0.428,\,0.534)$$ contains $$0.5$$.

Thus, there is not enough evidence to conclude that the therapists’ performance was different from chance.
**(a)** The expected proportion of correct responses by chance is 0.5. **(b)** The best point estimate of the therapists' success rate is 0.481. **(c)** The 95% confidence interval for the proportion of correct responses is approximately 0.428 to 0.534. **(d)** There is not enough evidence to suggest that the therapists performed better than chance.

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