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
![\alpha = 1 - CI](https://tex.z-dn.net/?f=%5Calpha%20%3D%201%20-%20CI)
![\alpha = 1 - 90\%](https://tex.z-dn.net/?f=%5Calpha%20%3D%201%20-%2090%5C%25)
![\alpha = 1 - 0.90](https://tex.z-dn.net/?f=%5Calpha%20%3D%201%20-%200.90)
![\alpha = 0.10](https://tex.z-dn.net/?f=%5Calpha%20%3D%200.10)
Divide by 2
![\frac{\alpha}{2} = 0.10/2](https://tex.z-dn.net/?f=%5Cfrac%7B%5Calpha%7D%7B2%7D%20%3D%200.10%2F2)
![\frac{\alpha}{2} = 0.05](https://tex.z-dn.net/?f=%5Cfrac%7B%5Calpha%7D%7B2%7D%20%3D%200.05)
Subtract from 1
![1 - \frac{\alpha}{2} = 1 - 0.05](https://tex.z-dn.net/?f=1%20-%20%5Cfrac%7B%5Calpha%7D%7B2%7D%20%3D%201%20-%200.05)
![1 - \frac{\alpha}{2} = 0.95](https://tex.z-dn.net/?f=1%20-%20%5Cfrac%7B%5Calpha%7D%7B2%7D%20%3D%200.95)
From the z table. the critical value for
is:
![z = 1.960](https://tex.z-dn.net/?f=z%20%3D%201.960)
Solving (c): Margin of error
This is calculated as:
![E = z * \frac{\sigma}{\sqrt n}](https://tex.z-dn.net/?f=E%20%3D%20z%20%2A%20%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%20n%7D)
![E = 1.960 * \frac{5}{\sqrt {150}}](https://tex.z-dn.net/?f=E%20%3D%201.960%20%2A%20%5Cfrac%7B5%7D%7B%5Csqrt%20%7B150%7D%7D)
![E = 1.960 * \frac{5}{12.25}](https://tex.z-dn.net/?f=E%20%3D%201.960%20%2A%20%5Cfrac%7B5%7D%7B12.25%7D)
![E = \frac{1.960 *5}{12.25}](https://tex.z-dn.net/?f=E%20%3D%20%20%5Cfrac%7B1.960%20%2A5%7D%7B12.25%7D)
![E = \frac{9.80}{12.25}](https://tex.z-dn.net/?f=E%20%3D%20%20%5Cfrac%7B9.80%7D%7B12.25%7D)
![E = 0.800](https://tex.z-dn.net/?f=E%20%3D%20%200.800)
Solving (d): The confidence interval
This is calculated as:
![CI = (\bar x - E, \bar x + E)](https://tex.z-dn.net/?f=CI%20%3D%20%28%5Cbar%20x%20-%20E%2C%20%5Cbar%20x%20%2B%20E%29)
![CI = (7.10 - 0.800, 7.10 + 0.800)](https://tex.z-dn.net/?f=CI%20%3D%20%287.10%20-%200.800%2C%207.10%20%2B%200.800%29)
![CI = (6.3, 7.9)](https://tex.z-dn.net/?f=CI%20%3D%20%286.3%2C%207.9%29)
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