Question

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

Answer: 1.467

Step-by-step explanation:

Formula of Margin of Error for (n<30):-

$E=t_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}$

Given : Sample size : n= 22

Level of confidence = 0.90

Significance level : $\alpha=1-0.90=0.10$

By using the t-distribution table ,

Critical value : $t_{n-1, \alpha/2}=t_{21,0.05}= 1.720743$

Standard deviation: $\sigma=4$

Then, we have

$E=(1.720743)\dfrac{4}{\sqrt{22}}=1.46745456106\approx1.467$

Hence, the margin of error for the confidence interval for the population mean with a 90% confidence level =1.467

Answer 2

Answer:

1.40

Step-by-step explanation:

SD =4

Sample Size (n)=22

z critical value for 90%CL = 1.645

Margin of Error = z*(sd/sqrt(n))

=1.645*(4/sqrt(22))

=1.645*0.8528

=1.40 (Rounded 2 decimal places)