A 90% confidence intervals for the proportion...

Question

A 90% confidence intervals for the proportion of first-year statistics students with brown hair is given to be (0.25, 0.41).

(a) What is the sample proportion of students with brown hair?

 (Round to 3 decimal places, if needed.)

(b) What is the margin of error of this interval?

 (Round to 3 decimal places, if needed.)

(c) What is the critical value used for this interval?

 (Round to 3 decimal places, if needed.)

(d) What is the standard error of the estimate?

 (Round to 3 decimal places, if needed.)

(e) What was the sample size?

 (Round to the appropriate integer.)

Answer 1

a) the sample proportion is the value that is exactly in the middle of the interval, so it is the average of the lower and upper bounds. (.25+.41)/2 = .33

b) The margin of error is the amount that you add and subtract to the sample proportion to get the upper and lower bounds, so it is .41-.33= .08 OR .33-.25=.08

c) With a 90% confidence interval, you can look it up on the t-table or in your graphing calculator use InvNorm(.95,0,1) = 1.645

d) The standard error is the margin of error / critical value= .08/1.645= .0486

e) The standard error for sample proportions= sqrt(p(1-p)/n). .0486 = sqrt(.33(1-.33)/n). Using algebra isolate the variable n. n= 93.6 rounded up to 94.