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A sample of 10 widgets has a mean of 21.950 and standard deviation of 6.350. At...

A sample of 10 widgets has a mean of 21.950 and standard deviation of 6.350. At 90% confidence, the lower limit with 3 decimal places is. Please show formulas and steps taken.

Math   Statistics-And-Probability   
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Answer #1

n = 10

= 21.950

s = 6.350

SE = s/

= 6.350/ = 2.0080

= 0.10

ndf = n - 1= 10 - 1 = 9

From Table, critical values of t = 1.8331

Confidence Interval:

21.950 (1.8331 X 2.0080)

= 21.950 3.6809

= (18.269 ,25.631)

So,

Lower limit = 18.269

So,

Answer is:

18.269

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