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igtguith
15.07.2020 •
Mathematics
You are testing the claim that the mean GPA of night students is different from the mean GPA of day students. You sample 30 night students, and the sample mean GPA is 2.35 with a standard deviation of 0.46. You sample 25 day students, and the sample mean GPA is 2.58 with a standard deviation of 0.47. Test the claim using a 5% level of significance. Assume the sample standard deviations are unequal and that GPAs are normally distributed. Hypotheses:
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Ответ:
Step-by-step explanation:
This is a test of 2 independent groups. The population standard deviations are not known. Let μ1 be the mean GPA of night students and μ2 be the mean GPA of day students.
The random variable is μ1 - μ2 = difference in the mean GPA of night students and the mean GPA of day students.
We would set up the hypothesis.
The null hypothesis is
H0 : μ1 = μ2 H0 : μ1 - μ2 = 0
The alternative hypothesis is
H1 : μ1 ≠ μ2 H1 : μ1 - μ2 ≠ 0
Since sample standard deviation is known, we would determine the test statistic by using the t test. The formula is
(x1 - x2)/√(s1²/n1 + s2²/n2)
From the information given,
x1 = 2.35
x2 = 2.58
s1 = 0.46
s2 = 0.47
n1 = 30
n2 = 25
t = (2.35 - 2.58)/√(0.46²/30 + 0.47²/25)
t = - 1.82
The formula for determining the degree of freedom is
df = [s1²/n1 + s2²/n2]²/(1/n1 - 1)(s1²/n1)² + (1/n2 - 1)(s2²/n2)²
df = [0.46²/30 + 0.47²/25]²/[(1/30 - 1)(0.46²/30)² + (1/25 - 1)(0.47²/25)²] = 0.00025247091/0.00000496862
df = 51
We would determine the probability value from the t test calculator. It becomes
p value = 0.075
Alpha = 5% = 0.05
Since alpha, 0.05 < than the p value, 0.075, then we would fail to reject the null hypothesis.
Ответ:
.0
014206323
Step-by-step explanation:
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