Question

A CI is desired for the true average stray-load loss μ (watts) for a certain type of induction motor when the line current is held at 10 amps for a speed of 1500 rpm. Assume that stray-load loss is normally distributed with σ = 2.2. (Round your answers to two decimal places.)

(a) Compute a 95% CI for μ when n = 25 and x = 53.8.

  ,

watts

(b) Compute a 95% CI for μ when n = 100 and x = 53.8.

  ,

watts

(c) Compute a 99% CI for μ when n = 100 and x = 53.8.

  ,

watts

(d) Compute an 82% CI for μ when n = 100 and x = 53.8.

,

watts

(e) How large must n be if the width of the 99% interval for μ is to be 1.0? (Round your answer up to the nearest whole number.)
n =

Answer 1

solution:-

given that σ = 2.2

(a) 95% CI for μ when n = 100 and x = 53.8

the value of 95% confidence from z table is 1.96

confidence interval formula

=> x +/- z * σ/sqrt(n)

=> 53.8 +/- 1.96*2.2/sqrt(25)

=> (52.94 , 54.66)

(b) 95% CI for μ when n = 100 and x = 53.8

the value of 95% confidence from z table is 1.96

confidence interval formula

=> x +/- z * σ/sqrt(n)

=> 53.8 +/- 1.96*2.2/sqrt(100)

=> (53.37 , 54.23)

(c) 99% CI for μ when n = 100 and x = 53.8

the value of 99% confidence from z table is 2.58

confidence interval formula

=> x +/- z * σ/sqrt(n)

=> 53.8 +/- 2.58*2.2/sqrt(100)

=> (53.23 , 54.37)

(d) 82% CI for μ when n = 100 and x = 53.8

the value of 82% confidence from z table is 1.34

confidence interval formula

=> x +/- z * σ/sqrt(n)

=> 53.8 +/- 1.34*2.2/sqrt(100)

=> (53.51 , 54.09)

(e) here given that width is 1.0 and

the value of 99% confidence from z table is 2.58

formula

=> n = (2*z*σ/e)^2

=> n = (2*2.58*2.2/1.0)^2

=> n = 129