Question

Suppose 240 randomly selected people are surveyed to determine whether or not they play a musical instrument. Of the 240 surveyed, 60 reported playing a musical instrument.

Answer 1

Answer:

[ 0.19522 < p^ < 0.30478 ]

Step-by-step explanation:

Solution:-

- A randomly selected sample of n = 240 people are surveyed to determine whether they play any instrument or not.

- The results for the survey showed that that 60 out of 240 people played a musical instrument.

- The proportion of people who played a musical instrument from the surveyed sample is:

                             p = 60 / 240 = 1/4

                             p = 0.25

- Where ( p ) is the sample proportion of the number of people who played any musical instrument.

- We are to construct a 95% confidence interval for the population proportion for which we are to determine the Margin of Error ( ME ).

- The general form is given for a confidence interval is:

                       [ p - ME < p^ < p + ME ]

Where,

                p^ : The population proportion

- The margin of error for a confidence interval is given as:

                    $ME = Z-critical* \sqrt{\frac{p*(1-p) }{n} }$

- The Z-critical value is defined by the Confidence level. So for Confidence level of 95% the Z-critical is determined by:

                   Confidence Level = 95%

                   Significance Level ( α ) = 1 - CI/100 = 1 - 95/100 = 0.05

                   Z-critical = Z_α/2 = Z_0.025

                   Z-critical = Z_0.025 = 1.96

- Therefore the margin of error would be:

                  $ME = 1.96* \sqrt{\frac{0.25*(1-0.25) }{240} }\\\\ME = 1.96* \sqrt{\frac{0.25*(0.75) }{240} }\\\\ME = 1.96* 0.02795\\\\ME =0.05478$

- The confidence can then be written as:

                  [ p - ME < p^ < p + ME ]

                  [ 0.25 - 0.05478 < p^ < 0.25 + 0.05478 ]

                  [ 0.19522 < p^ < 0.30478 ]