Question

In a survey of 650 community college students, 423 indicated that they have read a book for personal enjoyment during the school year. Construct a 90% confidence interval for the proportion of community college students who have read a book for personal enjoyment during the school year. Use 4 non-zero decimal places in your calculations.

a.Verify the normality condition and the independence condition

b.Find the Z-alpha/2

c.Find the margin of error (ME)

d. Find the upper/lower bound and Interpret the confidence interval

Answer 1

Out of 650 community college students, 423 indicated that they have read a book for personal enjoyment.

Hence random sample = n = 650, x = number of students read a book for personal enjoyment = 423

Hence proportion of students who read a book for personal enjoyment.

a. Conditions :

1. Here sample is random and independently drawn.

2.

Here , both are greater than 10. Hence here we use normal approximation.

Hence normality and indepenent conditions are satisfied.

b. Here confidence level is 0.90.

= 1- c = 1- 0.90 = 0.10,

We find z value corresponding to area 0.05 to the right of normal curve.

Total area under normal curve is 1. So remaining area to the left is 1-0.05 = 0.95

As standard normal probability table gives values for area to the left. Hence to find we use area 0.95.

= = 1.645 -------------( using excel formula " =norm.s.inv( 0.95)" )

c.

Margin of error is given by,

= 0.031

d. Hence the 90% confidence interval is given by,

- E < P < + E

0.65 - 0.031 < P < 0.65 + 0.031

0.619 < P < 0.681

Lower bound = 0.619

Upper bound = 0.681

Interpretation :

We are 90% confident that proportion of students who read a book for personal enjoyment lies between 0.619 and 0.681.