1 answer

5. Find the values or andズfor a sample of 25 scores and confidence interval of 98%....

Question:

5. Find the values or andズfor a sample of 25 scores and confidence interval of 98%. Drawズdiagram for this example.-イ! 6. The following are heights (in inches) of randomly selected women: 63.7, 61.2, 66.0,62.5,65.7,64.4,63.0, 61.9 a) Find the best point estimate of the population variance ơ. b) Construct a 95% confidenceinterva estimate ofthepopulationstandarddenationo. c) Does the confidence interval contain the standard deviation value of 2.5 inches? (Womens heights are known to have a standard deviation of 2.5 inches).
5. Find the values or andズfor a sample of 25 scores and confidence interval of 98%. Drawズdiagram for this example.-イ! 6. The following are heights (in inches) of randomly selected women: 63.7, 61.2, 66.0,62.5,65.7,64.4,63.0, 61.9 a) Find the best point estimate of the population variance ơ. b) Construct a 95% confidenceinterva estimate ofthepopulationstandarddenationo. c) Does the confidence interval contain the standard deviation value of 2.5 inches? (Women's heights are known to have a standard deviation of 2.5 inches).

Answers

5)

df= n-1 = 24

alpha=1-0.98 = 0.02

Low end alpha= 0.02/2 = 0.01

chi^2L =chi^20.01,24 = 42.9798

excel formula =CHIINV(0.01,24)

----------

high end alpha= 1-0.02/2 = 0.99

chi^2U=chi^20.99 = 10.8563

excel formula =CHIINV(0.99,24)

---------------------------------------------------------------------------------------------------------------------

6)

point estimate for population variance = sum [ x- mean]^2 /(n-1) = 3.003

b)

Confidence Interval Formula for σ is as follows: Square Root((n-1)s2/X2a2) < σ < Square Root(n-1)s2/X21-o2) where: (n-1-Degrees of Freedom. S2-sample variance and α 1-Confidence Percentage

alpha=0.05

alpha/2 = 0.025

sqrt (24*3.003/16.0128<sigma< sqrt (24*3.003/1.6899)

1.1457< sigma < 3.5269

so, confidence interval is (1.1457,3.5269)

c)

yes, confidence interval contain value 2.5

.

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