Respuesta :
Answer:
A) Null hypothesis; H0: μ1 − μ2 ≤ 0
Alternative hypothesis; Ha: μ1 − μ2 > 0
B) Test statistic = t = 1.878
C) p-value is 0.038823.
D) Reject the Null hypothesis H0
Step-by-step explanation:
We are given;
α = 0.05
n1 = 16
n2 = 10
bar x1 = 6.82
bar x2 = 6.28
s1 = 0.65
s2 = 0.75
A) The hypothesis is as follows;
Null hypothesis; H0: μ1 − μ2 ≤ 0
Alternative hypothesis; Ha: μ1 − μ2 > 0
B) Formula yo determine the test statistic is;
t = ((bar x1) - (bar x))/√((s1)²/n1) + (s2)²/n2))
Plugging in the relevant values, we have;
t = (6.82 - 6.28)/√((0.65)²/16) + (0.75)²/10))
t = 0.54/√(0.02640625 + 0.05625)
t = 0.54/0.2875
t = 1.878
C) The formula for the degree of freedom is;
Δ = [(s1)²/n1) + (s2)²/n2))]²/[((s1²/n1)²/(n1 - 1)) + ((s2²/n2)²/(n1 - 1))
Plugging in the relevant values, we have;
Δ = [(0.65)²/16) + (0.75)²/10))]²/[((0.65²/16)²/(16 - 1)) + ((0.75²/10)²/(10 - 1))
Δ = 0.00683205566/(0.000046486 + 0.0003515625)
Δ ≈ 17
Thus, the P-value;
From online p-value calculator from t-score and DF which i attached, we have the p-value as;
The p-value is 0.038823.
D) The p-value result is significant at p < 0.05
Thus, we reject the Null hypothesis H0

A) Null hypothesis: μ1 − μ2 ≤ 0
B) Test statistic = t = 1.878
C) The p-value is 0.038823.
D) Reject the Null hypothesis H0.
Hypothesis
What all information we have ?
α = 0.05
n1 = 16
n2 = 10
bar x1 = 6.82
bar x2 = 6.28
s1 = 0.65
s2 = 0.75
Part A)
The hypothesis is as follows;
Null hypothesis; H0: μ1 − μ2 ≤ 0
Alternative hypothesis; Ha: μ1 − μ2 > 0
Part B)
The formula to determine the test statistic is :
t = ((bar x1) - (bar x))/√((s1)²/n1) + (s2)²/n2))
t = (6.82 - 6.28)/√((0.65)²/16) + (0.75)²/10))
t = 0.54/√(0.02640625 + 0.05625)
t = 0.54/0.2875
t = 1.878
The formula to determine the test statistic is t = 1.878.
Part C)
The formula for the degree of freedom is;
Δ = [(s1)²/n1) + (s2)²/n2))]²/[((s1²/n1)²/(n1 - 1)) + ((s2²/n2)²/(n1 - 1))
Δ = [(0.65)²/16) + (0.75)²/10))]²/[((0.65²/16)²/(16 - 1)) + ((0.75²/10)²/(10 - 1))
Δ = 0.00683205566/(0.000046486 + 0.0003515625)
Δ ≈ 17
Thus, the P-value is 0.038823.
Part D)
The p-value result is significant at p < 0.05 is :
Thus, we reject the Null hypothesis H0.
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