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)

Sample standard deviation = [tex]s=\sqrt{\frac{(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}[/tex]

Sample standard deviation = [tex]s=7.149[/tex]

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,[tex]t_{\frac{\alpha}{2}}=1.833[/tex]

Formula of confidence interval = [tex]\bar{x}\pm t_{\frac{\alpha}{2}} \times \frac{s}{\sqrt{n}}[/tex]

Substitute the values

Confidence interval = [tex]57\pm 1.833 \times \frac{7.149}{\sqrt{10}}[/tex]

Confidence interval = [tex]57- 1.833 \times \frac{7.149}{\sqrt{10}}[/tex] to [tex]57+ 1.833 \times \frac{7.149}{\sqrt{10}}[/tex]

Confidence interval = [tex]52.856 [/tex] to [tex]61.143[/tex]

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