Question

A random survey of enrollment at 35 community colleges across the United States yielded the following figures: 6,415; 1,549; 2,107; 9,348; 21,829; 4,300; 5,943; 5,724; 2,824; 2,043; 5,483; 5,199; 5,855; 2,749; 10,010; 6,359; 27,002; 9,416; 7,679; 3,201; 17,502; 9,202; 7,380; 18,313; 6,558; 13,714; 17,769; 7,491; 2,769; 2,862; 1,262; 7,283; 28,163; 5,082; 11,624. Assume the underlying population is normal.

• Part (a)

  Find the following. (Round your answers to the nearest whole number.)(i)    

  x =  

• s_x =  

• n = 35

• n − 1 = 34

• Which distribution should you use for this problem? (Enter your answer in the form z or t_df where df is the degrees of freedom.)

• Part (d)

  Construct a 95% confidence interval for the population mean enrollment at community colleges in the United States.(i) State the confidence interval. (Round your answers to two decimal places.)

  ( , )

  
  
  (ii) Sketch the graph. (Round your answers to two decimal places. Enter your α/2 to three decimal places.)

  	CL = 0.95
  α 2 = 0.025		α 2 =0.025
  	0.05

  
  (iii) Calculate the error bound. (Round your answer to two decimal places.)

Answer 1

a) x = (6415 + 1549 + 2107 + 9348 + 21829 + 4300 + 5943 + 5724 + 2824 + 2043 + 5483 + 5199 + 5855 + 2749 + 10010 + 6359 + 27002 + 9416 + 7679 + 3201 + 17502 + 9202 + 7380 + 18313 + 6558 + 13714 + 17769 + 7491 + 2769 + 2862 + 1262 + 7283 + 28163 + 5082 + 11624)/35 = 8629

s_x = sqrt(((6415 - 8629)^2 + (1549 - 8629)^2 + (2107 - 8629)^2 + (9348 - 8629)^2 + (21829 - 8629)^2 + (4300 - 8629)^2 + (5943 - 8629)^2 + (5724 - 8629)^2 + (2824 - 8629)^2 + (2043 - 8629)^2 + (5483 - 8629)^2 + (5199 - 8629)^2 + (5855 - 8629)^2 + (2749 - 8629)^2 + (10010 - 8629)^2 + (6359 - 8629)^2 + (27002 - 8629)^2 + (9416 - 8629)^2 + (7679 - 8629)^2 + (3201 - 8629)^2 + (17502 - 8629)^2 + (9202 - 8629)^2 + (7380 - 8629)^2 + (18313 - 8629)^2 + (6558 - 8629)^2 + (13714 - 8629)^2 + (17769 - 8629)^2 + (7491 - 8629)^2 + (2769 - 8629)^2 + (2862 - 8629)^2 + (1262 - 8629)^2 + (7283 - 8629)^2 + (28163 - 8629)^2 + (5082 - 8629)^2 + (11624 - 8629)^2)/35) = 6844

We should use t-distribution.

d) df = 35 - 1 = 34

At 95% confidence interval the critical value is t_{0.025, 34} = 2.032

The 95% confidence interval for population mean is

+/- t_{0.025, 34} * s/

= 8629 +/- 2.032 * 6844/

= 8629 +/- 2350.71

= 6278.29, 10979.71

iii) Margin of error = t_{0.025, 34} * s/

                             = 2.032 * 6844/

                             = 2350.71