Question

Will rate!!

is desired for the true average stray ad loss μ (watts) for a certain type of hduction motor when the lire currert is h ld at 10 ames for age dd isoo mm Assume tat stra lad les is n distributed with σ-3.19. Round your answers to two decimal places. You must enter the LOWER BOUND first then the UPPER BOUND second.) (a) Compute a 95% CI for μ when n-25 and x-52.91. watts (b) Compute a 95% CI for μ when n-100 and x-52.01. (c) Compute a 99% ci for μ when n " 100 and x-52.91. watts width of the 99% interval for μ is to be 1-0? (Round you mwe wto the nearer d) How large must n be if the You may need to use the z table to complete this problem

Answer 1

a) Here i have used M for mean instead of X bar and also S instead of Sigma

The formula for estimation of confidence interval is:

μ = M ± t(s_M), Since n=25 which is less than 30 so t statistic is applicable here

where:

M = sample mean
t = t statistic determined by confidence level
s_M = standard error = √(s²/n)

Now,

M = 52.91
t = 2.06
s_M = √(3.19²/25) = 0.64

μ = M ± t(s_M)
μ = 52.91 ± 2.06*0.64
μ = 52.91 ± 1.3168

95% CI [51.59, 54.23].

b) here n=100 which is greater than 30 so Z statistic is applicable here

Also,

μ = M ± Z(s_M)

where:

M = sample mean
Z = Z statistic determined by confidence level
s_M = standard error = √(s²/n)

M = 52.91
Z= 1.96
s_M = √(3.19²/100) = 0.32

μ = M ± Z(s_M)
μ = 52.91 ± 1.96*0.32
μ = 52.91 ± 0.6252

95% CI [52.29, 53.54]

c) Again,

Calculation

M = 52.91
t = 2.58
s_M = √(3.19²/100) = 0.32

μ = M ± Z(s_M)
μ = 52.91 ± 2.58*0.32
μ = 52.91 ± 0.8217

99% CI [52.09, 53.73].

d)

The minimum sample is calculated by

$n=Z_{c}^{2}*\frac{\sigma^{2}}{E^{2}}$

where E=1

$n=2.58^{2}*\frac{3.19^{2}}{1^{2}}=67.73\approx 68$

The Z value is calculated using the table Shown below, This table shows the Z value in the left according to the P value in the middle