Question

A random sample of n1=130 individuals results in x1=35 successes. An independent sample of n2=149 individuals results in x2=59 successes. Does this represent sufficient evidence to conclude that p1<p2 at the ?=0.1 level of significance?

questions:

the given situation is about?

write the hypotheses for the test

h0:

h1:

calculate the test statistic:

(remaining in photo)

Idely the region See the cod choice below and finale bors within your choice (Type an integracimal rounded to two decimal places as needed) OA What is the thes and concludere CHEY WCS

Answer 1

Solution Given data X = 35 n =130 X = 59 ng = 149 x=0.1 Pi 35 130 Pi= 0.2692 X2 P2= ne 59 149 P2 = P₂ = 0.3960 P= Xi+X2 hitna P= 35+59 130t 149 94 Pa 271 P= 0.3369

Po state the hull and Now to alternative hypothesis Nute hypothesis Ho: Pi= P2 Altea native hypothesis #, : Pi<Pz This is "left-tailed" test Now to calculate the test statistic We know that The formula of test statistic, z is given by Test statistic, z = VG-P) ni ne 0.2692 – 0.3960 N 0.3369 (1-0.3369) + 130 149 -0.1268 z 0.3364 (0.6631) (0.00769+0.0067 -0.1268 z = Vous 0.3369 (0.6631) (0.0144) -0.1268 Z 0.00322

3 - 0.1268 z = 0.0567 Z = -2.24 Test statistic, Z=-2.24 Now to find the Z critical Value The Z-conitical value for a left-tailed for a significance level of &=o.is z = -1.28 Critical region : Test statistic < -1.28 Therefore the option-c" is the correct answer

Population is greater than the * Decision Test statistic, Z < Ecritical value - 2.24 < -1.28 since test statistic z is less than the Z critical value. So we do not reject the mall hypothesis (Ho) Conclusion : (do not reject the null hypothesis and conclude there is not sufficient evidence that the Parameter of interest of individual e at the 2=0.60 level of significance Population