We want to obtain a sample to estimate a population mean. Based on previous evidence, researchers believe the population standard deviation is approximately o = 27.9. We would like to be 99.5% confident that the estimate is within 10 of the true population mean. How large of a sample size is required? n=

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A: given data normal distributionμ = 241σ = 17n= 10

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A: Answer  By using the given data:  10 pounds the first two weeks with population standard deviation…

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A: given data claim : μ > 4.3 Mb/sn =39x¯ = 4.9 Mb/sσ = 1.1 Mb/s

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A: From given : n=42x¯=5.3s=18.5

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A: The sample size is 80, sample mean is 1150 and sample standard deviation is 150.

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A: From given :There are 31 student in bus School A = 8 School B = 31-8 =23

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A: Suppose the variable of interest is x.

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A: The mean is 100 and standard deviation is 15.

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A: I used Minitab for my calculations.

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A: Given that: Population mean, μ=165 Population standard deviation, σ=10 Size of sample, n=26

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A: P(defective shipment) = p = 0.01sample size (n) = 29accept the whole batch if there is at most one…

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A: Suppose a random variable x defines the contribution by each family.

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A: given data, p=0.15E=3%=0.03CI=0.98α=1-0.98=0.02n=?

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A: Given that.   X~N( μ , ?)   μ=5.2 , ?=0.7   Z-score =( x - μ )/?

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A: We have given that Mean(µ) = 100Standard deviations (σ) = 15X ~ N (µ, σ )= N(100, 15)

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Q: Find the confidence level related to the critical value za/2= 2.19 88.9% 97.1% 81.7% 97.9%

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A:

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A: Given The data is as follows: Sample proportion, p^=0.67

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Q: IQ scores are known to be normally distributed with a mean of 100 and a standard deviation of 15.…

A:

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A: It is given that a surveys 1000 accounts and finds that 91 are more than 30 days overdue.