Question

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 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​ Emily's hand without seeing it and without touching it. Among 329 ​trials, the touch therapists were correct 154 times.
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 random guesses?
b. Using Emily's sample results, what is the best point estimate of the therapists' success rate?
c. Using Emilys sample results, construct a 90% confidence interval estimate of the proportion of correct responses made by touch therapists.

Answer 1

Answer:

a) $p = 0.5$

b) $\hat p= \frac{X}{n}= \frac{154}{329}= 0.468$

c) $0.468 - 1.64 \sqrt{\frac{0.468(1-0.468)}{329}}=0.423$

$0.468 + 1.64 \sqrt{\frac{0.468(1-0.468)}{329}}=0.513$

And the 90% confidence interval for the true proportion is given by (0.527;0.593).

Step-by-step explanation:

Part a

For this case we don't have any prior info given so then the probability assumed for this case is:

$p = 0.5$

Part b

For this case the proportion of success is given by:

$\hat p= \frac{X}{n}= \frac{154}{329}= 0.468$

Part c

The confidence interval for the true proportion would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 90% confidence interval the value of and , and the critical value would be given by:

$z_{\alpha/2}=1.64$

The confidence interval is given by:

$0.468 - 1.64 \sqrt{\frac{0.468(1-0.468)}{329}}=0.423$

$0.468 + 1.64 \sqrt{\frac{0.468(1-0.468)}{329}}=0.513$

And the 90% confidence interval for the true proportion is given by (0.527;0.593).