Question
Among a random sample of 500 college students, the mean number of hours worked per week at non-college related jobs is 14.6. This mean lies 0.4 standard deviations below the mean of the sampling distribution. If a second sample of 500 students is selected, what is the probability that for the second sample, the mean number of hours worked will be less than 14.6?
A. 0.5
B. 0.6179
C. 0.6554
D. 0.3446
A. 0.5
B. 0.6179
C. 0.6554
D. 0.3446
Answers
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√n
If only one SD is provided, you can use just that to determine SEdiff.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability
related to the Z score.
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√n
If only one SD is provided, you can use just that to determine SEdiff.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability
related to the Z score.
o.5
b)0.6179
o.6554
0.3446
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