Answer:
Explanation:
This is a test of 2 independent groups. The population standard deviations are not known. it is a two-tailed test. Let c b the subscript for students using computer and n be the subscript for students not using computer.
Therefore, the population means would be μc and μn.
The random variable is xc - xn = difference in the sample mean scores of students who used computers and those who didn't use computers
We would set up the hypothesis.
The null hypothesis is
H0 : μc = μn H0 : μc - μn = 0
The alternative hypothesis is
Ha : μc ≠ μn Ha : μc - μn ≠ 0
Since sample standard deviation is known, we would determine the test statistic by using the t test. The formula is
(xc - xn)/√(sc²/nc + sn²/nn)
From the information given,
xc = 309
xn = 303
sc = 29
sn = 32
nc = 60
nn = 40
t = (309 - 303)/√(29²/60 + 32²/40)
t = 0.953
The formula for determining the degree of freedom is
df = [sc²/nc + sn²/nn]²/(1/nc - 1)(sc²/nc)² + (1/nn - 1)(sn²/nn)²
df = [29²/60 + 32²/40]²/(1/60 - 1)(29²/60)² + (1/40 - 1)(32²/40)² = 1569.48/20.13
df = 78
We would determine the probability value from the t test calculator. It becomes
p value = 0.344
Assuming a level of significance of 0.05, we would not reject the null hypothesis because the p value, 0.344 is > 0.05
Therefore, we cannot conclude that the population mean scores differ.