Question

1. What is proportion of the students exercise regularly? Select a random sample of 95 students and 62 said they exercise regularly. Construct 90% confidence interval for the proportion of all students exercising regularly.
   1. Find the critical value.
   2. Calculate the margin of error.
   3. Write the interval.

Please show work, thanks!

Answer 1

Solution :

Given that,

n = 95

x = 62

Point estimate = sample proportion = = x / n = 62/95=0.653

1 -   = 0.347

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05 = 1.645 ( Using z table )

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

= 1.645 *((0.653*0.347) / 95)

E = 0.080

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.653-0.080 < p < 0.653+0.080

0.573< p < 0.733