Answer:
(48.106 ; 53.494)
Step-by-step explanation:
Given the data:
X : 52 48 49 52 53
Sample mean = ΣX / n
n = 5
Sample mean, xbar = 254 / 5 = 50.8
Standard deviation, s = 2.17 (using calculator)
The standard error (SE) : s/√n =2.17/√5 = 0.970
The degree of freedom, df = n-1
df = 5 - 1 = 4
Tscore(0.05, 4) = 2.776
Confidence interval :
Xbar ± Tscore*standard error
50.8 ± (2.776 * 0.970)
50.8 ± 2.694
Lower boundary = 50.8 - 2.694 = 48.106
Upper boundary = 50.8 + 2.694 = 53.494
(48.106 ; 53.494)
Answer:
i dont know have a good day pls stay safe dont get hurt and get a good grade
Step-by-step explanation:
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