A bank offers auto loans to qualified customers. The amount of the loans are normally distributed and have a known population standard deviation of 4 thousand dollars and an unknown population mean. A random sample of 22 loans is taken and gives a sample mean of 42 thousand dollars. Find the margin of error for the confidence interval for the population mean with a 90% confidence level.

Answer: 1.467Step-by-step explanation:Formula of Margin of Error for (n<30):-[tex]E=t_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}[/tex]Given : Sample size : n= 22Level of confidence = 0.90Significance level : [tex]\alpha=1-0.90=0.10[/tex]By using the t-distribution table , Critical value : [tex]t_{n-1, \alpha/2}=t_{21,0.05}= 1.720743[/tex]Standard deviation: [tex]\sigma=4[/tex]Then, we have[tex]E=(1.720743)\dfrac{4}{\sqrt{22}}=1.46745456106\approx1.467[/tex]Hence, the margin of error for the confidence interval for the population mean with a 90% confidence level =1.467