Question

In: Statistics and Probability

QUESTION 1 1. Use the data below to answer the following questions: 11.5 12.9 13.5 15.2...

QUESTION 1 1. Use the data below to answer the following questions:

11.5

12.9

13.5

15.2

10.8

12.6

13.8

13.1

14.5

12.9

1. Calculate the sum of squares

2. Calculate the SEM.

3. Calculate a 95% confidence interval: Xbar +/- _________________. 1 points QUESTION 4 4. Calculate the 95% Lower Limit (Xbar - CI) 1 points

5. Calculate the 95% Upper Confidence Limit (Xbar + 95% CI)

6. You have a set of 15 observations of 10th rib pork backfat with a mean of 0.72 and s of 0.23. Construct a 95% IC for the true mean of 0.72 +/-___________________.

7. The upper 95% confidence limit is______________________.

8. The lower 95% confidence limit is ________________.

9. Calculate a 99% confidence interval with 0.72 +/- _____________________.

Expert Solution

	Values ( X )	Σ ( X_i- X̅ )²
	11.5	2.4964
	12.9	0.0324
	13.5	0.1764
	15.2	4.4944
	10.8	5.1984
	12.6	0.2304
	13.8	0.5184
	13.1	0.0004
	14.5	2.0164
	12.9	0.0324
Total	130.8	15.196

Mean X̅ = Σ Xi / n
X̅ = 130.8 / 10 = 13.08
Sample Standard deviation S_X = √ ( (Xi - X̅ )² / n - 1 )
S_X = √ ( 15.196 / 10 -1 ) = 1.2994

Sum of Squares = 15.196

SEM = S / √ n = 1.2994 / √ 10 = 0.4109

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.05 /2, 10- 1 ) = 2.262
13.08 ± t(0.05/2, 10 -1) * 1.2994/√(10)
Lower Limit = 13.08 - t(0.05/2, 10 -1) 1.2994/√(10)
Lower Limit = 12.1505
Upper Limit = 13.08 + t(0.05/2, 10 -1) 1.2994/√(10)
Upper Limit = 14.0095
95% Confidence interval is ( 12.1505 , 14.0095 )

Part 6)

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.05 /2, 15- 1 ) = 2.145
0.72 ± t(0.05/2, 15 -1) * 0.23/√(15)
Lower Limit = 0.72 - t(0.05/2, 15 -1) 0.23/√(15)
Lower Limit = 0.5926
Upper Limit = 0.72 + t(0.05/2, 15 -1) 0.23/√(15)
Upper Limit = 0.8474
95% Confidence interval is ( 0.5926 , 0.8474 )

Margin of Error = t(α/2, n-1) S/√(n) = 0.1274

0.72 +/- 0.1274

Lower Limit = 0.5926

Upper Limit = 0.8474

Part 9)

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.01 /2, 15- 1 ) = 2.977
0.72 ± t(0.01/2, 15 -1) * 0.23/√(15)
Lower Limit = 0.72 - t(0.01/2, 15 -1) 0.23/√(15)
Lower Limit = 0.5432
Upper Limit = 0.72 + t(0.01/2, 15 -1) 0.23/√(15)
Upper Limit = 0.8968
99% Confidence interval is ( 0.5432 , 0.8968 )

Margin of Error = t(α/2, n-1) S/√(n) = 0.1768

0.72 +/- 0.1768

orchestra answered 1 year ago

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