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? 8.75 Incorrect: Your answer is incorrect. (b) What is the point estimate of the population standard deviation? (Round your answer to three decimal places.) (c) With 95% confidence, what is the margin of error for the estimation of the population mean? (Round your answer to one decimal place.) (d) What is the 95% confidence interval for the population mean? (Round your answer to one decimal place.) to

Answers

Answer 1

Answer:

10 ; 3.464 ; 2.4 ; (7.6, 12.4)

Step-by-step explanation:

Given the data :

10, 8, 12, 15, 13, 11, 6, 5

Point estimate of population mean :

m = ΣX / n

n = sample size = 8

(10+8+12+15+13+11+6+5) / 8

= 10

Point estimate of population standard deviation :

Sqrt(Σ(X - m)^2 / n-1)

((10-10)^2 + (8-10)^2 + (12-10)^2 + (15-10)^2 + (13 - 10)^2 + (11-10)^2 + (6-10)^2 + (5-10)^2) / 7

= sqrt(84/7)

= 3.464

Margin of error at 95%:

Zcritical * sqrt(sd²/n)

Zcritical at 95% = 1.96

1.96 * sqrt(3.464^2 / 8)

Margin of Error = 2.4

Confidence interval :

m ± Z(s/sqrt(n))

10 - 1.96(3.4634/sqrt(8)) = 7.6

10 - 1.96(3.4634/sqrt(8)) = 12.4


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Answers

Answer:

(6x^(3)+11x^(2)-19x+5)-:-(3x-2)

Step-by-step explanation:[tex]63+112+32+1[/tex]

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121/2% in to a fraction ​

Answers

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9514 1404 393

Answer:

  1/8

Step-by-step explanation:

You can replace % with /100 to make it a fraction.

  12 1/2% = (12 1/2)/100 = (25/2)/100 = 25/200 = 1/8

__

Or, you can reduce the decimal fraction.

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