# In a science fair​ project, Emily conducted an experiment in which she tested pr

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 358 ​trials, the touch therapists were correct 172 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 random​ guesses? ​(Type an integer or a decimal. Do not​ round.)
b) Using​ Emily's sample​ results, what is the best point estimate of the​ therapists' success​ rate? ​(Round to three decimal places as​ needed.)
c) Using​ Emily's sample​ results, construct a $$\displaystyle{90}​\%$$ confidence interval estimate of the proportion of correct responses made by touch therapists.
Round to three decimal places as​ needed - ?$$\displaystyle{<}{p}{<}$$?

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Theodore Schwartz

Step 1
Given:
$$\displaystyle{n}={358}$$
$$\displaystyle{x}={172}$$
a) The proportion of correct responses is obtained as below:
$$\displaystyle\hat{{{p}}}={\frac{{{x}}}{{{n}}}}$$
$$\displaystyle={\frac{{{172}}}{{{358}}}}$$
$$\displaystyle={0.480}$$
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 $$\displaystyle{p}=\hat{{{p}}}$$
$$\displaystyle={0.480}$$
Thus, the best point estimate of the​ therapists' success​ rate is 0.480.
c) From the Standard Normal Table, the value of $$\displaystyle{z}\times$$ for $$\displaystyle{90}\%$$ level is $$\displaystyle{1.645}$$.
The 90​% confidence interval estimate of the proportion of correct responses made by touch therapist is obtained as below:
sample statistic $$\displaystyle\pm{z}\times{S}{E}=\hat{{{p}}}\pm{z}\times\sqrt{{{\frac{{\hat{{{p}}}{\left({1}-\hat{{{p}}}\right)}}}{{{n}}}}}}$$
$$\displaystyle={0.480}\pm{1.645}\times\sqrt{{{\frac{{{0.480}\times{0.520}}}{{{358}}}}}}$$
$$\displaystyle={0.480}\pm{\left({1.645}\times{0.0264}\right)}$$
$$\displaystyle={0.480}\pm{0.0434}$$
$$\displaystyle={\left({0.437},\ {0.523}\right)}$$
Thus, the $$\displaystyle{90}​\%$$ confidence interval estimate of the proportion of correct responses made by touch therapist is $$\displaystyle{0.437}{<}{p}{<}{0.523}$$.