Question 927869: The average household income per year is μ = $ 15,000. For a random sample of 15 households in the area, the mean x-bar = 14,000 with a sample standard deviation of s = $2,000. Test the hypothesized value of the mean at the 5 percent level of significance.
Answer by ewatrrr(24785)   (Show Source):
You can put this solution on YOUR website! Previously Posted
Ho: u = $15,000 (claim)
Ha: u ≠ $15,000
......
Sample: n = 15
mean x-bar = 14,000 with a sample standard deviation of s = $2,000
.......
Test statistic: (14000-15000)/(2000/sqrt(15)) = -1.9365
p-value: p( z < -1.9365) = normalcdf(-100, -1.9365) = .0264
..........
.05 level of significance....
.0264 < .05, Reject Ho.
The test results do NOT support the claim that u = $15,000
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