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The following sample data are from a normal population : 10,8,12,,15,13,11,6,5

User Yotka
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1 Answer

3 votes
Answer:
10
3.4641
2.8965
12.8965
Explanation:
Given: 10, 8, 12, 15, 13, 11, 6, 5
c = 95%
a. The point estimate of the population mean is the sample mean. The mean is the sum of all values divided by the number of values:
x = 10 + 8 + 12 + 15 + 13 + 11 + 6 + 5 /8
= 80/8
= 10
b. The point estimate of the population standard deviation is the sample standard deviation. The variance is the sum of squared deviations from the mean divided by n - 1. The standard deviation is the square root of the variance:
s = /(10 – 10)^2 +.... + (5– 10)^2/8 – 1
s = 3.4641
c. Determine the t-value by looking in the row starting with degrees of freedom df = n-1 = 8 –1 = 7 and in the column with = (1 – c)/2 = 0.025 in table :
t_/2 = 2.365
The margin of error is then:
E = t_/2 * s/√n
= 2.365 x s 3.4641/ √8
= 2.8965
d. The confidence intent)] then becomes:
7.1035 = 10 – 2.8965 = x – E
User Kbolino
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