Answer to Question #320243 in Statistics and Probability for John

Question

1. A researcher wants to compare the performance of students living with their parents and

the performance of these students whose parents are working. Random samples were

taken from two normally distributed groups of population. 20 students who are living

with their parents posted an average grade of 89.25 with standard deviation of 2.4,

whereas the 15 randomly selected students whose parents are working abroad have an

average grade of 86.00 with standard deviation of 3.2. Test whether there is a significant

difference between the two groups of students at 0.05 level of significance.

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DEADLINE : 04/1/2022

Answer 1

H₀:$\mu1=\mu2$

H₁:$\mu1 \not=\mu2$

n₁=20

sample mean 1 =89.25

s1=2.4

n2=15

sample mean 2 =86

s2=3.2

At 0.05 level of significance t critical value =1.98,-1.98

t=$\frac{sample mean 2-sample mean 1 }{\surd \frac{s1^2}{n1}+\frac{s2^2}{n2}}$

t=$\frac{3.2-2.4}{\surd \frac{2.4^2}{20}+\frac{3.2^2}{86 }}$

t=1.25

Since T statistics (1.25) lies between t critical value (-1.98 and 1.98) we fail to reject the null hypothesis. We conclude that there is insufficient evidence to conclude that there is a significant difference in performance between students living with their parents and students whose parents are working abroad.