Question

7.2 In a random sample of 400 students in a college of business, 176 favored the addition of more required mathematics courses in the freshman and sophomore years. In a random sample of 600 students in colleges of liberal arts, 216 favored the addition of more required mathematics courses in the freshman and sophomore years. a. Can you be reasonably sure (probability = 0.01) that the percentages of students in the two types of schools differed in their opinions on this subject? b. Construct a 93% confidence interval for the difference between the percentages.

Answer 1

ANSWER::

a)

null Hypothesis: Ho: alternate Hypothesis: Ha: P1P2 P1P2 0.0000 0.0000 for 0.01 level with two tailed test, critical value of z= 2.58 Decision rule: reject Ho if absolute value of test statistic [z>2.576 Business 176 payin, 0.4400 400 estimated difference in proportion =P1-P2 estimator of pooled proportion p =(x4+)(n+ng) Pa=xz/nz - na Liberal arts 216 0.3600 600 0.0800 0.3920 std error Se= (P:'(1-P) (1/+ 1/n) 0.0315 test statz (P1-P)/Se = 2.54

since test statistic does not falls in rejection region we fail to reject null hypothesis

we do not have have sufficient evidence to conclude that, there is a difference in percentage

b)

Business 176 Pi=xylny = 0.4400 400 estimated difference in proportion Pa=xz/nz = Liberal arts 216 0.3600 600 0.0800 =P1-P2 std error Se = (Pi" (1-P)/n+P2' (1-Palinz) 0.0316 for 93 Cl value of z= margin of error E-z' std error lower confidence bound= Upper confidence bounds 1.81 0.057 0.0227 0.1373

from above 93% confidence interval for difference in population proportion =(0.023,0.137)

OR

FOR BETTER UNDERSTANFING

© Here we have a P. Ryga 176 0044 Pe = Ryne - 216 - 0.36 OU 2 600 ² = x+82 - 176 +216 2, toe 400+600 = 0.392 0,01 Hypotheses Hoi P, = P2 Ha P &P Test statiste - P - Ý No2te) 0.44 -0.36 0.392 11-0.8 Y 0.39211-0.892) 1 Foottoo) -[254) pualue at 2=2.54 . pualne z 2112) 2.54) 22 [1- P1252.59) 7 -2c1-0,99453 270.0055/ since, pualue aooof, we regeet Hoi conclusion: There is sufficient evidence to support the claim and we conclude that shools differed in the is opinions ♡ At 93% confidence level de 1-0.93=0.07

Zerif = 70.07/2 = 1081 Nang 93 ci z = [0, -22) zert 18.(1=80) A de LIPPET 10.44 -0.367 I 1.81 x 9:4411–144) +36 114.36) -0.08 0.0572 92 4.00 boo 0.0228 001372

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