Question

In a random sample of 13 microwave​ ovens, the mean repair cost was ​$70.00 and the standard deviation was ​$15.40.

Using the standard normal distribution with the appropriate calculations for a standard deviation that is​ known, assume the population is normally​ distributed, find the margin of error and construct a 98​% confidence interval for the population mean. A 98​% confidence interval using the​ t-distribution was (58.5,81.5).

Find the margin of error of the population mean.

Find the confidence interval of the population mean.

Compare the results.

Answer 1

sample mean 'x̄=	70.000
sample size    n=	13.00
std deviation σ=	15.400
std error ='σx=σ/√n=	4.2712

for 98 % CI value of z=		2.33
margin of error E=z*std error =		9.936~ 9.9
lower bound=sample mean-E=		60.1
Upper bound=sample mean+E=		79.9
from above 98% confidence interval for population mean =(60.1 , 79.9)

from above we can see that margin of error and width of confidence interval with standard normal distribution is smaller in comparison when t distribution was used,