Submitted By lisiva

Words 400

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Words 400

Pages 2

A) By using the Central Limit Theorem with the sample being greater than 30 we can assume that the distribution and shape of x will be normally distributed.

B) Mean = 20 (no equations as it was given) 4/ 64 = 4/8 = 0.5 Standard Deviation

C) Z = (21-20) / (4/ 64) = (1/0.5) = 2 P = 1-.09772 = .0228

D) Z = (19.385 – 20) / (4/ 64) = -0.615/0.5 = -1.23 P = 0.1093

7.30

A) Z = (0.32 - 0.3)/√(0.3(1 - 0.3)/1011) = 1.3877 P = (p’ > 0.32) = P (z > 1.3877) = 0.0826

B) If we set the a to anything below 0.01 the claims can be refuted but on the other hand if we set it above than the claims would be true so the claims can neither be refuted or true .

8.8

A)

N = 100 X-bar = 5.46 s = 2.47 % = 95 Standard Error = σ/√n = 0.2470 z score = 1.9600

Width = 1.9600 * 0.2470 = 0.4841

Lower Limit = 5.46 – 0.4841 = 4.9759

Upper Limit = 5.46 + 0.4841 = 5.9441

Confidence interval of 95% [4.976, 5.944]

N = 100 X-bar = 5.46 s = 2.47 % = 99 Standard Error = σ/√n = 0.2470 z- score = 2.5758

Width = 2.5758 * 0.2470 = 0.6362

Lower Limit = 5.46 – 0.6362 = 4.8238

Upper Limit = 5.46 + 0.6362 = 6.0962

Confidence interval of 99% [4.824, 6.096]

B) Since the entire confidence interval is less than 6 the manager can confident that 95% is less than 6 minutes.

C) Because the confidence interval at 99% is over 6 the manager cannot be 99% confident that it’s less than 6 minutes.

D) If we choose α = 0.05, 95% is valid because it is less than 6 minutes, and not at 99% when it is α = 0.01.

8.38

N = 350 P = 0.557143 % = 95 Standard Error, SE = p(1 - p)/n} = 0.0266 z- score = 1.9600

Width = 1.9600 * 0.0266 = 0.0520

Lower Limit = .557143 – 0.0520 = 0.5051

Upper Limit = .557143 + 0.0520 = 0.6092

The confidence interval is [0.5051, 0.6092]

The answer is yes as the entire 95% of…...

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