Bus 308 Week 3 Assignments

In: Business and Management

Submitted By lisiva
Words 400
Pages 2
7.11
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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