n1=50n2=75x1ˉ=75s1=6x2ˉ=82s2=2H0:μ1=μ2H1:μ1≠μ2n_1 = 50 \\ n_2=75 \\ \bar{x_1}=75 \\ s_1=6 \\ \bar{x_2}= 82 \\ s_2=2 \\ H_0: \mu_1 = \mu_2 \\ H_1: \mu_1 ≠ \mu_2n1=50n2=75x1ˉ=75s1=6x2ˉ=82s2=2H0:μ1=μ2H1:μ1=μ2
Test-statistic:
Z=x1ˉ−x2ˉ(s12/n1)+(s22/n2)Z=75−82(62/50)+(22/75)=−7.96Z = \frac{\bar{x_1} -\bar{x_2}}{\sqrt{ (s_1^2/n_1)+(s^2_2/n_2) }} \\ Z= \frac{75-82}{\sqrt{ (6^2/50) +(2^2/75) }} = -7.96Z=(s12/n1)+(s22/n2)x1ˉ−x2ˉZ=(62/50)+(22/75)75−82=−7.96
Let use 5 % significant level.
Two-tailed test.
Reject H0 if Z≤ -1.96 or Z≥1.96.
Z=−7.96<Zcrit=−1.96Z= -7.96 < Z_{crit}= -1.96Z=−7.96<Zcrit=−1.96
Reject H0.
There is enough evidence to conclude that there is any difference between the performance of boys and girls at 5% significant level.
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