Answer:
(a) Point estimate = 7.10
(b) The critical value is 1.960
(c) Margin of error = 0.800
(d) Confidence Interval = (6.3, 7.9)
(e) We are 90% confident that the average number of hours worked by the students is between 6.3 and 7.9
Step-by-step explanation:
Given
-- sample mean
--- sample standard deviation
--- samples
Solving (a): The point estimate
The sample mean can be used as the point estimate.
Hence, the point estimate is 7.10
Solving (b): The critical value
We have:
--- the confidence interval
Calculate the
level




Divide by 2


Subtract from 1


From the z table. the critical value for
is:

Solving (c): Margin of error
This is calculated as:






Solving (d): The confidence interval
This is calculated as:



Solving (d): The conclusion
We are 90% confident that the average number of hours worked by the students is between 6.3 and 7.9