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)

Respuesta :

Answer:

Step-by-step 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

[tex]\bar x =6.77[/tex]

S.D = 0.21

[tex]I=6.77\pmt\times\frac{s}{\sqrt{n} }[/tex]

df = 24

α = 0.05

t = 2.064

[tex]I=6.77\pm2.064\times\frac{0.21}{\sqrt{25} } \\\\=6.77\pm0.087\\\\=[6.683,6.857][/tex]

b)

[tex]\sqrt{\frac{(1-n)s^2}{X^2_{\alpha /2} } < \mu <\sqrt{\frac{(1-n)s^2}{X^2_{1-\alpha/2} } }[/tex]

[tex]\sqrt{\frac{24 \times 0.21^2}{39.364} } < \mu <\sqrt{\frac{24 \times 0.21^2}{12.401} } \\\\=0.1640<\mu<0.2921[/tex]