Question

Let p equal the proportion of triathletes who suffered a training-related overuse injury during the past year. Out of 100 randomly selected triathletes, 80 indicated that they had suffered such an injury during the past year. Find an approximate 95% confidence interval for p

Answer 1

Solution :

Given that,

n = 100

x = 80

Point estimate = sample proportion = = x / n = 80/100=0.8

1 -   = 1- 0.8 =0.2

At 95% confidence level

  = 1 - 95% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z0.005 = 2.576 ( Using z table )

Margin of error = E = Z/2   * ((( * (1 - )) / n)

= 1.96 (((0.2*0.8) /100 )

= 0.0784

A 95% confidence interval is ,

- E < p < + E

0.2-0.0784 < p < 0.2+0.0784

0.1216< p < 0.2784