The following sample data are from a normal population: 10, 8, 12, 15, 13, 11, 6, 5.
a. What is the point estimate of the population mean?10 b. What is the point estimate of the population standard deviation (to 2 decimals)?3.46 c. With confidence, what is the margin of error for the estimation of the population mean (to 1 decimal)?2.9 d. What is the confidence interval for the population mean (to 1 decimal)?

10

3.4641

2.8965  

12.8965  

Step-by-step 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 <u<x +E= 10 + 2.8965 = 12.8965  

answer: asdfghjkl,mnbv

step-by-step explanation:

80x^8−68x^6+16^5−20x^4−4x^3+8x^2

hope i'm right lol