Question

One hundred draws are made at random with replacement from box A, and 250 are made at random with replacement from box B. (a) 50 of the draws from box A are positive, compared to 131 from box B: 50.0% versus 52.4%. Is this difference real, or due to chance

Answer 1

Answer:

Therefore WE accept the Null hypothesis $H_0$ That 50.0% versus 52.4%.difference is real

Step-by-step explanation:

Sample size A $n_a=100$

Sample size B $n_b=250$

Positive draw from box A $n_a=50$

Positive draw from box b $n_b=131$

Generally the equation for Probability of Positive draw from Box A is mathematically given by

 $P_1=\frac{50}{100}$

 $P_1=50%$

Therefore

 $1-P_1=50\%$

Generally the equation for Probability of Positive draw from Box A is mathematically given by

 $P_1=\frac{131}{250}$

 $P_1=52.4%$

Therefore

 $1-P_1=47.6\%$

Generally the equation for Standard error S.E is mathematically given by

 $S.E=\sqrt{\frac{n_1(p_1)*(1-P_1)}{n_1^2}+\frac{n_2(P_2(1-p^2))}{n_2^2}}$

 $S.E=\sqrt{\frac{100(0.50)*(50)}{100^2}+\frac{(250)(0.524(47))}{250^2}}$

 $S.E=0.592$

 $S.E=59.2\%$

Therefore

 $Z=\frac{50-52.4}{59.2}$

Generally

 $P value P>0.05$

Therefore WE accept the Null hypothesis $H_0$ That 50.0% versus 52.4%.difference is real