Based on 20 samples, the average resistance is 25 ohms and the sample standard deviation is 0.5 ohm. Determine the 90% confidence interval of the standard deviation of the batch. (10 points)

90% Confidence Interval = [24.81, 25.19]

Step-by-step explanation:

From the question, our number of samples is 20, it is small because it is less than 30. Hence we use the t score confidence interval formula

= Mean ± t score × Standard deviation/√n

Mean = 25 ohms

Standard deviation = 0.5 ohms

n = 20

We find the degrees of freedom = n - 1

= 20 - 1 = 19

Using the T score table

T score for 90% confidence interval with degrees of freedom 19

= 1.729

Hence:

Confidence Interval =

= 25 ± 1.729 × 0.5/√20

= 25 ± 0.1933080767

=

Confidence Interval

25 - 0.1933080767

= 24.806691923

≈ 24.81

25 + 0.1933080767

= 25.1933080767

≈ 25.19

90% Confidence Interval = [24.81, 25.19]

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