Question

Please show all work and provide legible replies.

1. In March of 2019, a (fictional) survey by your professor asked 78 medical workers if they enjoyed playing video games during their free time.  It was found that 11 of them responded positively.

Obtain an 85% confidence interval for the proportion of medical workers who enjoy playing video games during their free time

Answer 1

Sol)

Given

n = 78

x = 11

Phat = x / n = 11 / 78 = 0.1410

Qhat = 1 - phat = 1 - 0.1410 = 0.859

Z value for 85% confidence

Alpha = 1 - 0.85 = 0.15

Alpha / 2 = 0.15 / 2 = 0.075

Z value for alpha / 2 = 0.075 is 1.44

Confidence interval

= Phat +/- Z * SQRT ( Phat * qhat / n)

= 0.1410 +/- 1.44 ( SQRT ( 0.859 * 0.1410 ) / 78 ))

= 0.1410 +/- 1.44 ( SQRT ( 0.00155280))

= 0.1410 +/- 1.44 ( 0.03940 )

= 0.1410 +/- 0.056744

=( 0.1410 - 0.056744 , 0.1410 + 0.056744 )

= ( 0.084256 , 0.197744)