Question

# In a random sample of 13 microwave​ ovens, the mean repair cost was ​\$70.00 and the...

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.

 sample mean 'x̄= 70 sample size    n= 13 std deviation σ= 15.4 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,