Answer:
The 98% confidence interval for the mean time needed to complete this step is between 52.0403 seconds and 57.1597 seconds
Explanation:
We have that to find our
level, that is the subtraction of 1 by the confidence interval divided by 2. So:
![\alpha = (1-0.98)/(2) = 0.01](https://img.qammunity.org/2021/formulas/mathematics/college/1jmpexvujebtbpcrhdufd2s3v69fkvyi0a.png)
Now, we have to find z in the Ztable as such z has a pvalue of
.
So it is z with a pvalue of
, so
![z = 2.327](https://img.qammunity.org/2021/formulas/mathematics/college/ycihe6ulpjrwlumf0nrtee2h17bsc6wh4e.png)
Now, find the margin of error M as such
![M = z*(\sigma)/(√(n))](https://img.qammunity.org/2021/formulas/mathematics/college/cvh8tdoppqkhyobio78yaazk1nqj1870w9.png)
In which
is the standard deviation of the population and n is the size of the sample.
![M = 2.327(11)/(√(100)) = 2.5597](https://img.qammunity.org/2021/formulas/mathematics/college/znvopea144ob8h1zl29ig99if6cfa5hc1v.png)
The lower end of the interval is the sample mean subtracted by M. So it is 54.6 - 2.5597 = 52.0403 seconds
The upper end of the interval is the sample mean added to M. So it is 54.6 + 2.5597 = 57.1597 seconds
The 98% confidence interval for the mean time needed to complete this step is between 52.0403 seconds and 57.1597 seconds