A researcher wishes to see if the mean number of days that a basic, low price, small automobile sits on a dealer’s lot is 29. A sample of 39 automobile dealers has a mean of 32.31 days for basic, low price, small automobiles. At α=0.05, test the claim that the mean time is greater than 29 days. The standard deviation of the population is 4.26 days. What is the test value?

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A:

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A: μ  = 506 , σ = 100  Z~ N(0,1)

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A: Given Information Now we have to calculate corresponding Z value                         Z = 0.7388…

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A: Test for Standard deviation: The χ2 Chi-Square test is used to test the significance difference…

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A: Answer:-

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Q: If the scores on a test have a mean of 43 and standard deviation of 1.5, what is z score of 45?

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A: The mean is 496 and the standard deviation is 67.

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A: Given Data Population Mean,μ = 53.0 Population Standard Deviation,σ = 6.0

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A: Given: x~N(μ,σ)μ=95.6σ=24.3

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A:

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A: