The American Management Association is studying the income of store managers in the retail industry. A random sample of 49 managers reveals a sample mean of $45,420. The standard deviation of the population is known to be $2,050. a). Compute a 95% confidence interval, as well as the margin of error.
b). Interpret the confidence interval you have computed.

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.

explanation:

step-by-step explanation:

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