Question

An engineering consulting firm wantedto evaluate a rivet process by measuring the formed diameter. The following data represent the diameters (in hundredths of an inch) for a random sample of 24 rivet heads:

6.81 - 6.79 - 6.69 - 6.59 - 6.65 - 6.60 - 6.74 - 6.70 - 6.76
6.84 - 6.81 - 6.71 - 6.66 - 6.76 - 6.76 - 6.77 - 6.72 - 6.68
7.71 - 6.79 - 6.72 - 6.72 - 6.72 - 6.79 - 6.83
a) Set up a 95% confidence interval estimate of the average diameter of rivet heads (in hundredths of an inch).
b) Set up a 95% confidence interval estimate of the standard deviation of the diameter of rivet heads (in hundredths of an inch)

Answer 1

Explanation:

6.81 - 6.79 - 6.69 - 6.59 - 6.65 - 6.60 - 6.74 - 6.70 - 6.76

6.84 - 6.81 - 6.71 - 6.66 - 6.76 - 6.76 - 6.77 - 6.72 - 6.68

7.71 - 6.79 - 6.72 - 6.72 - 6.72 - 6.79 - 6.83

`\bar x =6.77`

S.D = 0.21

`I=6.77\pmt*(s)/(√(n) )`

df = 24

α = 0.05

t = 2.064

`I=6.77\pm2.064*(0.21)/(√(25) ) \\\\=6.77\pm0.087\\\\=[6.683,6.857]`

b)

`\sqrt{((1-n)s^2)/(X^2_(\alpha /2) ) < \mu <\sqrt{((1-n)s^2)/(X^2_(1-\alpha/2) ) }`

`\sqrt{(24 * 0.21^2)/(39.364) } < \mu <\sqrt{(24 * 0.21^2)/(12.401) } \\\\=0.1640<\mu<0.2921`