7s^3-9s^2+s+6 need help
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Answer:
a) The 95% confidence interval for the income of store managers in the retail industry is ($44,846, $45,994), having a margin of error of $574.
b) The interval mean that we are 95% sure that the true mean income of all store managers in the retail industry is between $44,846 and $45,994.
Step-by-step explanation:
Question a:
We have to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So:
Now, we have to find z in the Z-table as such z has a p-value of .
That is z with a p-value of , so Z = 1.96.
Now, find the margin of error M as such
In which is the standard deviation of the population and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 45420 - 574 = $44,846.
The upper end of the interval is the sample mean added to M. So it is 45420 + 574 = $45,994.
The 95% confidence interval for the income of store managers in the retail industry is ($44,846, $45,994), having a margin of error of $574.
Question b:
The interval mean that we are 95% sure that the true mean income of all store managers in the retail industry is between $44,846 and $45,994.