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

Anna flipped a coin to choose either her right hand or her left hand, and also she asked the therapists to identify the named hand by placing their hand just under Anna's hand without seeing it and without touching it. Among 358 trials, the touch therapists were correct 172 times. Complete parts​ (a) through​ (d).
a) Given that Anna used a coin toss to elect either her right hand or her left hand, what proportion of correct responses would be anticipated if the touch therapists made arbitrary suppositions?( Type an integer or a numeric. Do not round.)
b) Using Anna's sample results, what's the stylish point estimate of the therapists' success rate?( Round to three decimal places as demanded.)
c) Using Anna's sample results, construct a $90\mathrm{%}$ confidence interval estimate of the proportion of correct responses made by touch therapists.
Round to three decimal places as demanded-?$?

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Step 1
Given:
$n=358$
$x=172$
a) The proportion of correct responses is obtained as below:
$\stackrel{^}{p}=\frac{x}{n}$
$=\frac{172}{358}$
$=0.480$
Thus, the proportion of correct responses is 0.480.
Step 2
b) The best point estimate of the​ therapists' success​ rate is obtained as below:
Point estimate of $p=\stackrel{^}{p}$
$=0.480$
Thus, the best point estimate of the​ therapists' success​ rate is 0.480.
c) From the Standard Normal Table, the value of $z×$ for $90\mathrm{%}$ level is $1.645$.
The 90​% confidence interval estimate of the proportion of correct responses made by touch therapist is obtained as below:
sample statistic $±z×SE=\stackrel{^}{p}±z×\sqrt{\frac{\stackrel{^}{p}\left(1-\stackrel{^}{p}\right)}{n}}$
$=0.480±1.645×\sqrt{\frac{0.480×0.520}{358}}$
$=0.480±\left(1.645×0.0264\right)$
$=0.480±0.0434$

Thus, the $90\mathrm{%}$ confidence interval estimate of the proportion of correct responses made by touch therapist is $0.437.