Question

Ten engineers working for start-up companies were asked how long they worked, in hours, per week. The data (in hours) are 70, 45, 55, 60, 65, 55, 55, 60, 50, 55. Construct a 90% confidence interval for the population average length of time, in hours, for all engineers at start- ups work per week. (in 2 decimal)

Answer 1

`52.86` to
`61.14`

Explanation:

Data : 70, 45, 55, 60, 65, 55, 55, 60, 50, 55.

Sample mean =
`\bar{x}=\frac{\sum{x}}{n}`

Sample mean =
`\bar{x}=(70+45+55+60+65+55+55+60+50+55)/(10)`

Sample mean =
`\bar{x}=57`

Sample standard deviation =
`s=\sqrt{\frac{\sum(x-\bra{x})^2}{n-1}`

Sample standard deviation =
`s=\sqrt{((70-57)^2+(45-57)^2+(55-57)^2+(60-57)^2+(65-57)^2+(55-57)^2+(55-57)^2+(60-57)^2+(50-57)^2+(55-57)^2)/(10-1)`

Sample standard deviation =
`s=7.149`

Since n < 30 and population standard deviation is unknown

So we will use t dist.

Confidence interval = 90%

So, significance level =α= 0.1

Degree of freedom = n-1 = 10-1 =9

Now refer the t dist table for t critical

So,
`t_{(\alpha)/(2)}=1.833`

Formula of confidence interval =
`\bar{x}\pm t_{(\alpha)/(2)} * (s)/(√(n))`

Substitute the values

Confidence interval =
`57\pm 1.833 * (7.149)/(√(10))`

Confidence interval =
`57- 1.833 * (7.149)/(√(10))` to
`57+ 1.833 * (7.149)/(√(10))`

Confidence interval =
`52.856` to
`61.143`

Hence a 90% confidence interval for the population average length of time, in hours, for all engineers at start- ups work per week is
`52.86` to
`61.14`