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 296 ​trials, the touch therapists were correct 143 times.
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?

Answer 1

Answer:

a 0.5

b 0.4831

c 0.4354 < P < 0.53008

Step-by-step explanation:

Given that :

Probability (P) of a head or a tail when a coin is being tossed or flipped = 1/2 = 0.5

Sample size (n) = 296

Selected sample (X) = 143

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?

The  proportion of correct responses that would be expected if the touch therapists made random guesses is 0.5

b) Using Emily's sample results, what is the best point estimate of the therapists' success rate?

Point estimate $(p') = \frac{X}{n}$

= $(p') = \frac{143}{296}$

= 0.4831

c) Using Emily's sample results, construct a 90% confidence interval estimate of the proportion of correct responses made by touch therapists.

The $Z_c$ for 90% is 1.645

Using the formula P" -E < P < P" + E

where E = margin of error :  $Z_c * \sqrt{\frac{P(1-P)}{n} }$

$=1.645 * \sqrt{\frac{0.4831(0.5169)}{296} }$

$=1.645 * \sqrt{\frac{0.2497}{296} }$

$=1.645*0.029$

= 0.0477

∴ P" -E < P < P" + E

= 0.4831 - 0.0477 < P < 0.4831 + 0.0477

= 0.4354 < P < 0.53008