A sample of 40 transistors yielded a sample average saturation current of 9.46 milliamps, with a...

A sample of 40 transistors yielded a sample average saturation current of 9.46 milliamps, with a sample standard deviation of 0.58 milliamps. Give a 95% confidence interval for the true mean saturation current of this type of transistor. Show all work.

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Expert Solution

Solution :

Given that,

= 9.46

s =0.58

n = Degrees of freedom = df = n - 1 =40 - 1 = 39

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2= 0.05 / 2 = 0.025

t /2,df = t0.025,39 = 2.023 ( using student t table)

Margin of error = E = t/2,df * (s / $\sqrt$ n)

= 2.023 * (0.58 / $\sqrt$ 40)

= 0.19

The 95% confidence interval is,

- E < < + E

9.46 - 0.19 < <9.46 + 0.19

9.27< < 9.65

orchestra answered 1 year ago

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